NCERT Exemplar Q & A

Hint

(d) Key concept: In this situation if the particle is thrown in \(x\)-y plane (as shown in figure) at some angle \(\theta\) with velocity v , then we have to resolve the velocity of the particle in rectangular components, such that one component is along the field \((\mathrm{v} \cos \theta)\) and other one is perpendicular to the field \((\mathrm{v} \sin \theta)\). We find that the particle moves with constant velocity \(\mathrm{v} \cos \theta\) along the field. The distance covered by the particle along the magnetic field is called pitch.

The pitch of the helix, (i.e., linear distance travelled in one rotation) will be given by
\(
p=T(v \cos \theta)=2 p \frac{m}{q B}(v \cos \theta)
\)
For given pitch \(p\) correspond to charge particle, we have
\(
\frac{q}{m .}=\frac{2 \pi v \cos \theta}{q B}=\text { constant }
\)
Here in this case, charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic field B, LHS for two particles should be same and of opposite sign. Therefore,
\(
\left(\frac{e}{m}\right)_{1}+\left(\frac{e}{m}\right)_2=0
\)

Hint

(a) According to the Biot-Savart law, the magnitude of \(\vec{B}\) is: \(B \propto|q| ; B \propto v\);
\(
\begin{array}{r}
B \propto \sin \phi ; B \propto \frac{1}{r^2} \\
B \propto \frac{|q| v \sin \phi}{r^2} \\
B=\frac{\mu_0}{4 \pi} \frac{|q| v \sin \phi}{r^2}
\end{array}
\)
where \(\mu_0 / 4 \pi\) is a proportionality constant, ‘ \(r\) ‘ is the magnitude of position vectur from charge to that point at which we have to find the magnetic field and \(\phi\) is the angle between \(\vec{v}\) and \(\vec{r}\).
or \(\vec{B}=\frac{\mu_0}{4 \pi} \frac{|q|(\vec{v} \times \vec{r})}{\left|r^3\right|} \hat{n}\)
Where \(\hat{n}\) is the direction of \(\vec{B}\) which is in the direction of cross product of \(\vec{v}\) and \(\vec{r}\). Or we can say that \(\vec{B} \perp\) to both \(\vec{v}\) and \(\vec{r}\).

Hint

(a) Key concept: Direction of magnetic moment ( \(M=14\) ) of circular loop (in figure (a)) is perpendicular to the loop by right hand thumb rule. So to compare these magnetic moments, we have to analyse them vectorically. Now let us first analyse the situation:

The direction of magnetic moment of circular loop of radius R is placed in the x -y plane is along z -direction and given by \(M=\ln R\), (as shown in above figure-a). When half of the loop with \(x>0\) is now bent so that it now lies in the \(y\)-z plane as shown in the figure below.

The magnitudes of magnetic moment of each semicircular loop of radius \(R\) lie in the \(x-y\) plane and the \(y-z\) plane is \(M_1=M_2=I \frac{\pi R^2}{2}\) and the direction of magnetic moments are along \(z\)-direction and \(x\)-direction respectively. Their resultant
\(
M_{\mathrm{net}}=\sqrt{M_1^2+M_2^2}=\sqrt{2 I} \frac{\pi R^2}{2}=\frac{M}{\sqrt{2}}
\)
So, \(M_{\text {net }}<M\) or \(M\) diminishes.

Hint

(d) Step 1: Understand the setup
We have a long solenoid carrying a current, which produces a magnetic field inside it. An electron is projected with a uniform velocity along the axis of this solenoid.
Step 2: Determine the direction of the magnetic field
Inside a long solenoid, the magnetic field \((B)\) is directed along the axis of the solenoid. According to the right-hand rule, if you curl the fingers of your right hand in the direction of the current, your thumb points in the direction of the magnetic field.
Step 3: Analyze the motion of the electron
The electron is moving along the axis of the solenoid, which means its velocity \((v)\) is also directed along the same axis.
Step 4: Apply the Lorentz force law
The force \((F)\) experienced by a charged particle moving in a magnetic field is given by the Lorentz force law:
\(
F=q(\mathbf{v} \times \mathbf{B})
\)
where \(\boldsymbol{q}\) is the charge of the electron, \(\mathbf{v}\) is the velocity vector, and \(\mathbf{B}\) is the magnetic field vector.
Step 5: Evaluate the cross product
Since the velocity of the electron is parallel to the magnetic field (both are along the axis of the solenoid), the angle between \(\mathbf{v}\) and \(\mathbf{B}\) is 0 degrees. The sine of 0 degrees is 0 , which means:
\(
F=q(\mathbf{v} \times \mathbf{B})=0
\)
Thus, the force acting on the electron is zero.
Step 6: Conclusion
Since there is no net force acting on the electron, it will continue to move with uniform velocity along the axis of the solenoid. Therefore, the correct answer is that the electron will continue to move with uniform velocity along the axis of the solenoid.
Final Answer
The electron will continue to move with uniform velocity along the axis of the solenoid.

Hint

(a) Cyclotron is a device used to accelerate positively charged particles (like a-particles, deutrons etc.)
It is based on the fact that the electric field accelerates a charged particle and the perpendicular magnetic field keeps it revolving in circular orbits of constant frequency. Thus a small potential difference would impart enormously large velocities if the particle is made to traverse the potential difference a number of times.

Hint

(d) As, work done,
\(
W=M B\left(\cos \theta_1-\cos \theta_2\right)=0
\)
(As there no change in angle between \(M\) and \(B\) and when loop in rotated by \(30^{\circ}\) )

Hint

(b) Angular momentum of electron \((\mathrm{L})={nh} / 2 \pi\)
Magnetic moment of electron \(\mathrm{M}={neh} / 4 \pi m\)
gyromagnetic ratio \(=\) magnetic moment/angular momentum
\(2 \pi n e h / 4 \pi n h m\)
\(={e} / 2 {~m}\)
\(=\) constant
It is independent of the orbit in which electron is revolving.
Since e is negative, the geomagnetic ratio is negative. verifies the answer is (b)

Hint

(b, d) Key concept: If a current carrying straight conductor (length I) is placed in a uniform magnetic field (B) such that it makes an angle \(\theta\) with the direction of field, then force experienced by it is \(\mathrm{F}_{\max }={BIl} \sin \theta\). Direction of this force is obtained by right-hand palm rule.
Right-hand palm rule: Stretch the fingers and thumb of right hand at right angles to each other. Then if the fingers point in the direction of field \(B\) and thumb in the direction of current \(z\), then normal to the palm will point in the direction of force.
If conductor is placed perpendicular to magnetic field, then \(\theta=90^{\circ}, F_{\max }=BIl\)

Hint

(b, c) Key concept: Ampere’s law gives another method to calculate the magnetic field due to a given current distribution.

Line integral of the magnetic field \(\vec{B}\) around any closed curve is equal to \(\mu_0\) times the net current \(i\) threading through the area enclosed by the curve, i.e.
\(
\oint \vec{B} \cdot \overrightarrow{d I}=\mu_0 \Sigma i=\mu_0\left(i_1+i_3-i_2\right)
\)
Total current crossing the above area is \(\left(i_1+i_3-i_2\right)\). Any current outside the area is not included in net current. (Outward \(\odot \rightarrow+\mathrm{ve}\), Inward \(\otimes \rightarrow-\mathrm{ve}\) )
Applying the Ampere’s circuital law, we have \(\oint \mathbf{B} \cdot \mathbf{d} \mathbf{l}=\mathrm{i}_0(I-I)=0\) (because current is in opposite sense.)
Also, there may be a point on \(C\) where \(\mathbf{B}\) and \(\mathbf{d l}\) are perpendicular and hence,
\(
\oint \mathbf{B} \cdot \mathbf{d} \mathbf{l}=0
\)

Hint

(b, c, d) This problem is based upon the single moving charge placed with some uniform electric and magnetic fields in space. Then they experience a force called Lorentz force given by the relation \(F_{\text {net }}=q E+q(v \times B)\).
i. The magnetic forces \(\left(F_m=q(v \times B)\right)\), on charge particle is either zero or \(F_m\) is perpendicular to \(v\) (or component of \(v\) ) which in turn revolves particles on a circular path with uniform speed. In both cases, particles have equal accelerations.
ii. Due to the same electric force \(\left(\mathrm{F}_{\mathrm{e}}=\mathrm{qE}\right)\) which is in opposite direction (because of the sign of charge) both the particles gain or loss energy at the same rate.
iii. There is no change of the Centre of Mass (CM) of the particles, therefore the motion of the Centre of Mass (CM) is determined by B alone.

Hint

(a, b, d) This problem is based upon the single moving charge placed with some uniform electric and magnetic fields in space. Then they experience a force called Lorentz force is given by the relation \(\mathrm{F}_{\text {net }}=\mathrm{qE}+\mathrm{q}(\mathrm{v} \times \mathrm{B})\).

Force experience by the charged particle due to electric field \(\mathrm{F}_{\mathrm{e}}=\mathrm{qE}\)

Force experience by the charged particle due to magnetic field, \(\mathrm{F}_{\mathrm{m}}=\mathrm{q}(\mathrm{v} \times \mathrm{B})\)
According to the problem, particle is moving with constant velocity means the acceleration of particle is zero and also it is not changing its direction of motion.
This will happen when the net force on particle is zero.
i. if \(E=0\), and \(v \| B\), then \(F_{\text {net }}=0\).
ii. if \(\mathrm{E} \neq 0, \mathrm{~B} \neq 0\) and \(\mathrm{E}, \mathrm{v}\) and B are mutually perpendicular.
iii. When both E and B are absent.

Hint

(d) A positively-charged particle projected towards east can be considered as current in the eastern direction. Here, the positive charge is deflected towards the north by a magnetic field, i.e. the positively-charged particle experiences a force in the northern direction.

Hence, in order to determine the direction of the magnetic field, we apply Fleming’s left-hand rule. According to this rule, when we stretch the thumb, the fore-finger and the middle finger mutually perpendicular to each other, then the thumb gives the direction of the force experienced by the charged particle, the fore-finger gives the direction of the magnetic field and the middle finger gives the direction of the current. Thus, if we direct the middle finger in the eastern direction, the thumb in the northern direction, we see that the fore-finger points in the downward direction.

Thus, the direction of the magnetic field is found to be in the downward direction.

Hint

(d) The tension in the string may either increase or decrease depending on the direction of the magnetic force relative to the centripetal force.

When the charged particle is whirled in a horizontal circle, at any moment, the current direction can be taken along the tangent of the circle. Also, the magnetic field is in the vertical direction. So, using Fleming’s left-hand rule, the force can be radially outward or inward, depending on the direction of the magnetic field, i.e. either upward or downward. Also, the direction of force depends on the direction of the whirl, i.e. clockwise or anticlockwise and obviously on the charge of the particle, i.e. whether it is positive or negative. So, the correct answer is that the tension may increase or decrease.

Hint

(d) The magnetic force on a moving charge is given by the formula \(F=q v \boldsymbol{B} \sin (\theta)\), where \(q\) is the charge of the particle, \(v\) is its velocity, \(B\) is the magnetic field strength, and \(\theta\) is the angle between the velocity and the magnetic field. Since the particles are projected perpendicularly to the magnetic field, \(\theta=90^{\circ}\) and \(\sin (\theta)=1\). Therefore, the force is directly proportional to the charge. Here all particles are given same velocity and \(B\) is same. So the particle with greater charge experience greater magnetic force. Among the given \(\mathrm{LI}^{++}\)has maximum charge so it experience greater force.

Hint

(a) We know that,
radius of circular path \((R)=\frac{m v}{q B}\)
Here, \(v, q\) and \(B\) are same.
So, \(R \propto m\).
The mass of electron is minimum. So, electron will describe the smallest circle when projected with same velocity perpendicular to magnetic field.

Hint

(d) Since, we know,
Frequency, \(f=\frac{q B}{2 \pi m} v ; \propto \frac{q}{m}\)
In the given particles,
\(\frac{q}{m}\) is a minimum for \(L i^{+}\).
\(\therefore L i^{+}\)will have the minimum frequency.

Hint

(a)

\(
\begin{aligned}
&\text { Circular loop is placed perpendicular to the magnetic field. Therefore, angle between area vector and magnetic field is zero. }\\
&\therefore \tau=I A B \sin 0^{\circ}=0
\end{aligned}
\)

Hint

(c) When a beam of protons and electrons, moving at the same speed, passes through a magnetic field perpendicular to the beam, they will be deviated by different angles and hence will separate. This is because the force on a charged particle in a magnetic field is proportional to the charge and the speed of the particle. Since electrons and protons have the same charge magnitude but opposite signs, they will experience magnetic forces in opposite directions. Additionally, the mass of a proton is much greater than that of an electron. Therefore, the deviation angles of the protons and electrons will be different, causing them to separate.

Hint

(c) The path of the charged particle will be a helix with uniform pitch. When a charged particle enters a uniform magnetic field at an acute angle, its velocity vector can be resolved into two components: one parallel to the field and one perpendicular to it. The component parallel to the field causes the particle to move along the field with constant velocity, while the perpendicular component causes it to move in a circle. The combination of these two motions results in a helix.

Hint

(d) The path of the particle will be a helix with nonuniform pitch. Here’s why:

Magnetic Field:
When a charged particle moves perpendicular to a uniform magnetic field, it experiences a magnetic force that causes it to move in a circular path.

Electric Field:
If the particle is also experiencing a force from an electric field (which is parallel to the magnetic field), this force will cause it to accelerate in the direction of the electric field.

Combined Effect:
The combined effect of the circular motion due to the magnetic field and the accelerated motion due to the electric field results in a helical path. This path will have a non-uniform pitch because the electric field will cause the velocity along the direction of the electric field to change, altering the pitch of the helix.

Hint

(d) zero

Explanation: When the current enters and leaves the circular wire at diametrically opposite points, the magnetic field generated by each half of the wire at the center cancels each other out. Therefore, the net magnetic field at the center of the wire is zero. Since the magnetic force on a charged particle is given by \(F=q v B\), where \(q\) is the charge, \(v\) is the velocity, and \(B\) is the magnetic field, and \(B\) is zero at the center, the magnetic force on the particle is also zero.

Hint

(b, d) The electric field can apply force on a charge in a state of rest or motion, but the magnetic field cannot apply force on a charge at rest. Hence, options (b) and (d) are correct.

Hint

(a, d) As the charged particle is at rest, its velocity, \({V}=0\) and magnetic force, \(\mathrm{F}={qVB}=0\). Hence, we cannot determine whether a magnetic field is present or not. But as the particle at rest experiences an electromagnetic force, the electric field must be non-zero. As electric force acts on a particle, whether it is at rest or in motion, an electric force must be present.

Hint

(c, d) As the particle gets deflected, a force acts on the particle. So, either it has got deflected due to the magnetic force or electric force; so, both the fields cannot be zero. Also, the particle can be deflected under the combined effect of magnetic and electric forces; so, both fields can be non-zero.

Hint

(a, b, d) A charged particle is moving in a gravity-free space.
The velocity of the particle is constant, which means there is no net force acting on it (Newton’s first law).
Analyzing Each Option:
Option (a): Electric field \((E)=0\) and Magnetic field \((B) \neq 0\).
If \(E=0\), the electric force ( \(F_{-} e=Q\) \times \(E\) ) is also 0.
The magnetic force ( \(F_{-} m=Q \times V \times B\) ) can be \(0\) if the angle between \(V\) and \(B\) is 0 (i.e., they are parallel).
Therefore, if both forces are zero, the net force is zero, and this option is possible.
Option (b): Electric field \((E)=0\) and Magnetic field \((B) \neq 0\).
If \({E}=0\), then \(F_e\) \(=0\).
The magnetic force can also be zero if the velocity is parallel to the magnetic field ( \(\theta=0\) ).
Thus, the net force can still be zero, making this option possible.
Option (c): Electric field \((E) \neq 0\) and Magnetic field \((B)=0\).
If \(E \neq 0\), then \(F_e\) \(=[latex]Q \times\) \(E\) is not zero.
With no magnetic field to counteract this force, the net force cannot be zero.
Therefore, this option is not possible.
Option (d): Electric field \(({E}) \neq 0\) and Magnetic field \(({B}) \neq 0\).
Here, both forces are present: \(F_{-} e=Q \times E\) and \(F \_m=Q \times V \times B\).
It is possible for these two forces to be equal in magnitude and opposite in direction, resulting in a net force of zero.
Therefore, this option is also possible.

Hint

(b) The electric field exerts a force \(q E\) on the charged particle, which always accelerates (increases the speed) the particle.
The particle can never be rotated in a circle by the electric field because then the radius of the orbit will keep on increasing due to the acceleration, which is not possible. So, options (c) and (d) are incorrect. On the other hand, a magnetic field does not change the magnitude of the velocity but changes only the direction of the velocity. Since the particle is moving in a circle, where its speed remains constant and only the direction of velocity changes, so it can only be achieved if \({E}=0\) and \(B \neq 0\).

Hint

(a, b) A charged particle experiences two types of forces when it moves through electric and magnetic fields:
The electric force \(\left(F_E\right)\) given by \(F_E=q E\), where \(q\) is the charge of the particle and \(E\) is the electric field.
The magnetic force \(\left(F_B\right)\) given by \(F_B=q(\mathbf{v} \times \mathbf{B})\), where \(\mathbf{v}\) is the velocity of the particle and \(\mathbf{B}\) is the magnetic field.
Condition for Undeflected Motion:
For the particle to go undeflected, the net force acting on it must be zero. This means that the electric force must balance the magnetic force:
\(
F_E+F_B=0
\)
Thus, we can write:
\(
q E+q(\mathbf{v} \times \mathbf{B})=0
\)
Analyzing the Magnetic Force:
The magnetic force depends on the angle between the velocity of the charged particle and the magnetic field. If the particle’s velocity is parallel to the magnetic field, the angle \(\theta\) between \(\mathbf{v}\) and \(\mathbf{B}\) is \(0^{\circ}\). Therefore, the sine component becomes:
\(
F_B=q v B \sin \left(0^{\circ}\right)=0
\)
This means that the magnetic force is zero when the velocity is parallel to the magnetic field.
Electric Force Acting Alone:
Since the magnetic force is zero, the only force acting on the particle is the electric force:
\(
F_E=q E
\)
If the electric field is directed in the same direction as the velocity of the charged particle, the particle will continue to move undeflected.
Conclusion:
It is indeed possible for a charged particle to go undeflected in a region with both electric and magnetic fields if the magnetic field is aligned with the velocity of the particle (making the magnetic force zero) and the electric field is in the same direction as the particle’s motion.

Hint

(a, b) If a charged particle goes unaccelerated in a region containing electric and magnetic fields,
\(
|q \vec{E}|=|q \vec{v} \times \vec{B}| \Rightarrow|\vec{E}|=|\vec{v} \times \vec{B}| \Rightarrow E=v B \sin \theta .
\)
Hence \(\vec{E}\) must be perpendicular to the plane containing \(\vec{V}\) and \(\vec{B}\).

Hint

(b, d) 

\(
\begin{aligned}
&\begin{aligned}
& \mathrm{Bq} \v=\frac{\mathrm{m}^2}{\mathrm{r}} \Rightarrow \mathrm{r}=\frac{\mathrm{m} v}{\mathrm{~Bq}} \\
& \Rightarrow \mathrm{r} \propto \frac{1}{\mathrm{q}}, \frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{\mathrm{q}_2}{\mathrm{q}_1}=\frac{2 \mathrm{e}}{\mathrm{e}}=2 \Rightarrow \mathrm{r}_1=2 \mathrm{r}_2
\end{aligned}\\
&\text { As they are projected from the same place, the circular touch each other. }
\end{aligned}
\)

Hint

(a, b) Use right-hand rule or \(\vec{F}=q \vec{v} \times \vec{B}\) for explanation.

If we apply a magnetic field along the y-direction, the circular path in \(x-z\) plane, the particle can move parallel to the magnetic direction. If we apply the magnetic field in the z-direction, the plane of the circular path will be \(x-y\) plane.

Hint

(b, c) Electric force due to a charged particle is \(q E\) and magnetic force is \(q VB\).
We can sort out the two wrong equations using dimensional analysis. Now, equating the above two forces. we get:
\(
\mathrm{E}=\mathrm{VB}
\)
Hence, analysing the answers using dimensional analysis, we see that the second term on the RHS of the equations (b) and (c) are not dimensionally correct. Thus, the options (b) and (c) are wrong.

Hint

(c) 

\(
\begin{aligned}
&\begin{aligned}
& \vec{F}=q \vec{V} \times \vec{B} \\
& =-e V \hat{i} \times B(-\hat{j}) \\
& \vec{F}=e V B \hat{k}
\end{aligned}\\
&\text { Force is vertically upwards perpendicular to plane of paper. }
\end{aligned}
\)

Note: A vertical wire is carrying current in upward direction, so the magnetic field produced will be anticlockwise (according to the right-hand thumb rule). As the electron beam is sent horizontally towards the wire, the direction of the current will be horizontally away from the wire (direction of conventional current is opposite to the direction of the negative charge). According to Fleming’s left-hand rule, the force will act in upward direction, deflecting the beam in the same direction.

Hint

(c) The magnetic force on a wire carrying an electric current \(i\) is given as \(\vec{F}=i \cdot(\vec{l} \times \vec{B})\), where \(l\) is the length of the wire and \(B\) is the magnetic field acting on it. If a current-carrying straight wire is kept along the axis of a circular loop carrying current, then according to the right-hand thumb rule, the magnetic field due to the wire on the current-carrying loop will be along its circumference, which contains a current element \(i d \vec{l}\).
\(
\text { The force on the current carrying conductor is } F=i \overrightarrow{d l} \times \vec{B}
\)
As angle between \(d l[latex] and [latex]B\) is zero.
So, the cross product will be
\(
\begin{aligned}
& (\vec{l} \times \vec{B})=0 \\
& \Rightarrow \vec{F}=0
\end{aligned}
\)
Thus, the straight wire will not exert any force on the loop.

Hint

(a) A proton beam is going from north to south, so the direction of the current due to the beam will also be from north to south. Also, an electron beam is going from south to north, so the direction of the current due to the beam will also be from north to south. The direction of conventional current is along the direction of the flow of the positive charge and opposite to the flow of the negative charge. The magnetic field generated due to them will enter the plane of paper in the west and come out of the plane of paper in the east, according to the righthand thumb rule. Since both the beams have currents in the same direction, they will apply equal and opposite forces on each other and, hence, will attract each other. Thus, the electron beam will be deflected towards the proton beam.

Hint

(d)

North south direction means the loop is lying in the vertical plane. By using the right hand thumb rule, the direction of the magnetic field can be found both at Point A & Point B. Upon its application we find that at both the points the direction of the magnetic field is towards the west.

Hint

(b) The force acting on the wire per unit length of a current-carrying wire having a current \(i_1\), due to another wire which is also carrying current \(i_2\) at a distance \(d\) can be written as,
\(
F=\frac{\mu_0 i_1 i_2}{2 \pi d}
\)
This can be applied in each face of the loop.
The force on the surface \(A B\) can be found to be as,
\(
F_{A B}=\frac{\mu_0 i_1 i_2}{2 \pi d}
\)
This force will be acting towards the wire.
Now let us find the force on the surface BC which can be written as,
\(
F_{C D}=\frac{\mu_0 i_1 i_2}{2 \pi(d+a)}
\)
This will be away from the wire. As the force is inversely proportional to distance, we can see that the force of the surface CD is less than force on the surface AB , we can write that,
\(
\left.\vec{F}_{A B}\right\rangle \vec{F}_{C D}
\)
From this we can understand that the resultant force will be towards the wire.
\(
\left|\vec{F}_{B C}\right|=\left|\vec{F}_{D A}\right|
\)
As they are at equal distance from the wire carrying current \(i_1\). Even though their direction will be opposite. Therefore we can write that,
\(
\vec{F}_{B C}=-\vec{F}_{A D}
\)
Therefore, the net force will be towards the wire carrying current \(i_1\).

Hint

(d) The force on a charged particle \(q\) moving with velocity \(v\) in a magnetic field \(B\) is given by
\(
\vec{F}=q(\vec{v} \times \vec{B})
\)
As the charge is moving along the magnetic line of force, the velocity and magnetic field vectors will point in the same direction, making a cross product.
\(
\begin{aligned}
& (\vec{v} \times \vec{B})=0 \\
& \Rightarrow \vec{F}=0
\end{aligned}
\)
So, the magnetic force on the particle will be zero.

Hint

(c) The magnetic force \((F)\) acting on the charged particle can be expressed using the formula:
\(
F=Q(V \times B)
\)
Since V is perpendicular to B , the magnitude of the magnetic force is:
\(
F=Q V B
\)
Centripetal Force:
The particle will move in a circular path due to the magnetic force acting as the centripetal force. The centripetal force required for circular motion is given by:
\(
F_c=\frac{m v^2}{R}
\)
Where \(m\) is the mass of the particle and \(R\) is the radius of the circular path.
Equating Forces:
Setting the magnetic force equal to the centripetal force, we have:
\(
Q V B=\frac{m v^2}{R}
\)
Solving for the Radius (R):
Rearranging the equation to solve for \(R[latex] gives:
[latex]
R=\frac{m v}{Q B}
\)
Calculating the Area (A):
The area \((A)\) of the circular path traced by the particle is given by:
\(
A=\pi R^2
\)
Substituting the expression for R :
\(
A=\pi\left(\frac{m v}{Q B}\right)^2
\)
\(
A=\pi \frac{m^2 v^2}{Q^2 B^2}
\)
From the expression for area, we can see that:
\(
A \propto m^2 v^2
\)
This indicates that the area is proportional to the square of the velocity of the particle.
The area bounded by the path described by the particle is proportional to the square of its velocity, which is also related to the kinetic energy of the particle (since kinetic energy \(K E \propto \frac{1}{2} m v^2\) ).
Therefore, the area is proportional to the kinetic energy of the particle.

Hint

(c) The kinetic energy gained by a particle accelerated through a potential difference \(V\) is \(q V\).
The centripetal force on a particle moving in a circle of radius \(R\) in a magnetic field \(B\) is \(q v B\).
The kinetic energy of a particle is \(\frac{1}{2} m v^2\).
Equate the kinetic energy gained by the particles to the kinetic energy of the particles.
\(
\begin{aligned}
q V & =\frac{1}{2} m v^2 \\
v & =\sqrt{\frac{2 q V}{m}}
\end{aligned}
\)
Equate the magnetic force to the centripetal force.
\(
\begin{aligned}
& q v B=\frac{m v^2}{R} \\
& v=\frac{q B R}{m}
\end{aligned}
\)
Equate the two expressions for \(v\) and solve for \(m\).
\(
\begin{aligned}
& \sqrt{\frac{2 q V}{m}}=\frac{q B R}{m} \\
& \frac{2 q V}{m}=\frac{q^2 B^2 R^2}{m^2} \\
& m=\frac{q B^2 R^2}{2 V}
\end{aligned}
\)
Find the ratio of the masses.
\(
\begin{aligned}
\frac{m_X}{m_Y} & =\frac{\frac{q B^2 R_1^2}{2 V}}{\frac{q B^2 R_2^2}{2 V}} \\
\frac{m_X}{m_Y} & =\frac{R_1^2}{R_2^2} \\
\frac{m_X}{m_Y} & =\left(\frac{R_1}{R_2}\right)^2
\end{aligned}
\)
The ratio of the mass of \(X\) to that of \(Y\) is \(\left(\frac{R_1}{R_2}\right)^2\).

Hint

(b) Step 1: Identify the direction of the magnetic fields created by the 20 A and 40 A currents.
Using the right-hand rule, the magnetic field due to the 20 A current will circulate clockwise around the wire, and the magnetic field due to the 40 A current will circulate counterclockwise around the wire.

Step 2: Calculate the magnetic field at the midpoint between the two wires.
The magnetic field due to a long straight current-carrying wire at a distance r from the wire is given by \(B=\frac{\mu_0 I}{2 \pi r}\).

Step 3: Since the wire is placed midway, the distance from each wire to the midpoint is the same. Let this distance be d .
The magnetic field at the midpoint due to the 20 A wire is \(B_1=\frac{\mu_0-20}{2 \pi d}\) and due to the 40 A wire is \(B_2=\frac{\mu_0-40}{2 \pi d}\).

Step 4: Determine the net magnetic field at the midpoint. Since the currents are in opposite directions, their magnetic fields at the midpoint will add up.
The net magnetic field \(B_{\text {net }}=B_2-B_1=\frac{\mu_0 \cdot 40}{2 \pi d}-\frac{\mu_0 \cdot 20}{2 \pi d}=\frac{\mu_0 \cdot 20}{2 \pi d}\).

Step 5: Determine the direction of the force on the wire carrying the antiparallel current.
The force on a currentcarrying wire in a magnetic field is given by \(F=I L B\). Since the current in the wire is antiparallel to the 20 A current, the force will be towards the wire carrying the 40 A current.

The magnetic force on the wire will be towards 40 A .

Hint

(c) The magnetic field due to a long straight wire is \(B=\frac{\mu_0 i}{2 \pi r}\).
The magnetic field is a vector quantity.
The magnetic field due to multiple wires is the vector sum of the individual fields.
Define the magnetic field due to each wire.
\(
\begin{aligned}
& B_1=\frac{\mu_0 i_1}{2 \pi r} \\
& B_2=\frac{\mu_0 i_2}{2 \pi r}
\end{aligned}
\)
\(r\) is the distance from the wire to the midpoint.
Set up the equation for the magnetic field when the currents are in the same direction.
\(
\begin{aligned}
& B_{\text {same }}=B_1-B_2=10 \mu \mathrm{~T} \\
& \frac{\mu_0 i_1}{2 \pi r}-\frac{\mu_0 i_2}{2 \pi r}=10 \mu \mathrm{~T} \\
& \frac{\mu_0}{2 \pi r}\left(i_1-i_2\right)=10 \mu \mathrm{~T}
\end{aligned}
\)
Set up the equation for the magnetic field when the direction of \(i_2\) is reversed.
\(
\begin{aligned}
& B_{\text {reversed }}=B_1+B_2=30 \mu \mathrm{~T} \\
& \frac{\mu_0 i_1}{2 \pi r}+\frac{\mu_0 i_2}{2 \pi r}=30 \mu \mathrm{~T} \\
& \frac{\mu_0}{2 \pi r}\left(i_1+i_2\right)=30 \mu \mathrm{~T}
\end{aligned}
\)
Divide the second equation by the first equation.
\(
\begin{aligned}
& \frac{\frac{\mu_0}{2 \pi r}\left(i_1+i_2\right)}{\frac{\mu_0}{2 \pi r}\left(i_1-i_2\right)}=\frac{30}{10} \\
& \frac{i_1+i_2}{i_1-i_2}=3
\end{aligned}
\)
Solve for the ratio \(i_1 / i_2\).
\(
\begin{aligned}
& i_1+i_2=3\left(i_1-i_2\right) \\
& i_1+i_2=3 i_1-3 i_2 \\
& 4 i_2=2 i_1 \\
& \frac{i_1}{i_2}=\frac{4}{2}=2
\end{aligned}
\)
The ratio of the currents \(i_1 / i_2\) is 2.

Hint

(a) The magnetic field due to a long straight wire is given by \(B=\frac{\mu_0 i}{2 \pi r}\).
The drift velocity of electrons is given by \(v_d=\frac{i}{n A e}\).
The magnetic field is a non-relativistic phenomenon.
The magnetic field due to a long straight wire is given by:
\(B=\frac{\mu_0 i}{2 \pi r}\)
The observer is moving with the drift velocity of the electrons. The magnetic field is a non-relativistic phenomenon. The magnetic field does not depend on the observer’s velocity.
The magnetic field seen by the observer is the same as the magnetic field produced by the wire.
\(
B=\frac{\mu_0 i}{2 \pi r}
\)
The magnetic field seen by the observer is \(\frac{\mu_0 i}{2 \pi r}\).

Hint

(c, d) 1. The Biot-Savart Law states that the magnetic field \(\vec{B}\) at a point in space due to a current element \(I \overrightarrow{d l}\) is given by:
\(
d \vec{B}=\frac{\mu_0 I}{4 \pi} \frac{d \vec{l} \times \vec{r}}{r^3}
\)
where \(\vec{r}\) is the position vector from the current element to the point where the magnetic field is being calculated, and \(r\) is the magnitude of \(\vec{r}\).
2. Positioning the Current Element:
In this scenario, the current element \(I \vec{d} \vec{d}\) is located at position \(\vec{r}\) and we want to find the magnetic field at the origin.
3. Applying the Biot-Savart Law:
Since we are calculating the magnetic field at the origin due to the current element at position \(\vec{r}\), we can substitute into the Biot-Savart Law:
\(
d \vec{B}=\frac{\mu_0 I}{4 \pi} \frac{d \vec{l} \times \vec{r}}{r^3}
\)
4. Considering the Direction:
The direction of the magnetic field due to a current element follows the righthand rule. The vector \(d \vec{l}\) is in the direction of the current, and \(\vec{r}\) points from the current element to the origin.
5. Evaluating the Options:
(d): \(-\frac{\mu_0 I}{4 \pi} \frac{d \vec{l} \times \vec{r}}{r^3}\) is correct as it includes the negative sign indicating the direction of the magnetic field.Option
(c): \(\frac{\mu_0 I}{4 \pi} \frac{\vec{r} \times d \vec{l}}{r^3}\) is also correct because \(\vec{r} \times d \vec{l}=-d \vec{l} \times \vec{r}\) (due to the properties of the cross product).

Hint

(a, b, c) Lorentz Force:
\(
\begin{aligned}
& q v B=q E \\
& \Rightarrow \text { Dimensions of } \mathrm{x}=[v]=\left[\frac{E}{B}\right]=\left[L T^{-1}\right] \\
& y=\frac{1}{\sqrt{\mu_o \epsilon_o}}=\sqrt{\frac{4 \pi}{\mu_o} \times \frac{1}{4 \pi \epsilon_o}}=\sqrt{\frac{9 \times 10^9}{10^{-7}}}=3 \times 10^8=c \\
& \Rightarrow \text { Dimensions of } \mathrm{y}=[c]=\left[L T^{-1}\right]
\end{aligned}
\)
Time constant of RC circuit \(=\mathrm{RC}\) so dimensionally \([\mathrm{RC}]=[\mathrm{T}]\)
\(
\Rightarrow z=\left[\frac{l}{R C}\right] \Rightarrow[z]=\left[L T^{-1}\right]
\)
Therefore, \(\mathrm{x}, \mathrm{y}\) and z have the same dimensions.

Hint

(b, c ,d)  The magnetic field \(B\) at a point near a current-carrying conductor is
\(
B=\frac{\mu_0 I}{2 \pi d} \dots(1)
\)
Here, \(\mu_0\) is magnetic permeability of free space, \(I\) is current in the conductor and \(d\) is the distance of the point from the conductor.

We have given that the direction of the current in a wire is along the Z-axis.
From equation (1), we can conclude that the magnetic field at a point near the current carrying wire depends only on the distance of that point from the wire as the current carrying from the wire is the same for all the points.

Let us consider points A and B on the same side of the current carrying wire at different distances from the wire. The magnitude of the magnetic field at points \(A\) and \(B\) is not the same as their distances from the wire are different. But according to right hand thumb rule, the directions of magnetic fields at these points \(A\) and \(B\) are the same in the upward direction. Therefore, we can have two points in the \(X-Y\) plane with same directions of the magnetic field but not the points for the magnetic field with the same magnitude and direction. Hence the option \(A\) is incorrect and the option B is correct.

Now let us consider two points B and C on the either sides of the wire at the same distance from the wire.The distance of the points B and C being the same from the wire, the magnitude of the magnetic field at points B and C is the same. Therefore, we can have two points in the \(\mathrm{X}-\mathrm{Y}\) plane with the same magnitude of the magnetic field.Hence, the option C is correct.

According to the right hand thumb rule, the direction of the magnetic field at point \(B\) is in upward direction and magnetic field at point \(C\) is in the downward direction. Therefore, we can have two points in the \(X-Y\) plane with different directions of the magnetic field.

Hint

(b, c) The magnetic field around a long, straight wire with a uniform current density is strongest at the surface of the wire. The magnetic field strength increases linearly from the center of the wire to the surface, where it reaches its maximum. The magnetic field is zero at the center of the wire, and beyond the surface, it decreases with distance from the wire.

The magnetic field around the wire is made up of concentric circles. The magnitude of the magnetic field is proportional to the current and inversely proportional to the distance from the wire.

Hint

(b, c) A hollow tube is carrying uniform electric current along its length, so the current enclosed inside the tube is zero.
According to Ampere’s law,
\(
\oint \vec{B} \cdot d \vec{l}=\mu_o i_{\text {inside }}
\)
Inside the tube ,
\(
\begin{aligned}
& \oint \vec{B} \cdot d \vec{l}=0, r<R \\
& \Rightarrow B_{\text {inside }}=\text { Constant } \\
& \Rightarrow B_{\text {axis }}=0
\end{aligned}
\)
The magnetic fields from points on the circular surface will point in opposite directions and cancel each other.
Outside the tube,
\(
\begin{aligned}
& B \times 2 \pi r=\mu_o i \\
& \Rightarrow B_{\text {outside }}=\frac{\mu_o i}{2 \pi r}, r>R
\end{aligned}
\)

Hint

(a) We have a coaxial cable consisting of a central conductor and an outer conductor. The central conductor carries a current \(I\) in one direction, while the outer conductor carries an equal current \(I\) in the opposite direction.
Apply Ampere’s Law:
According to Ampere’s Law, the magnetic field \(B\) around a current-carrying conductor can be calculated using the formula:
\(
\oint \mathbf{B} \cdot d \mathbf{l}=\mu_0 I_{\mathrm{enc}}
\)
Here, \(I_{\text {enc }}\) is the current enclosed by the Amperian loop.
Choose an Amperian Loop:
We can consider different Amperian loops: one inside the inner conductor, one in the space between the inner and outer conductors, and one outside the outer conductor.
Evaluate the Magnetic Field in Different Regions:
Inside the Inner Conductor:
The magnetic field \(B\) is zero because there is no current enclosed by the loop.
Between the Inner and Outer Conductor:
The enclosed current is \(I-I=0\) (since the currents are equal and opposite), thus:
\(
B=0
\)
Outside the Outer Conductor:
The enclosed current is also zero because the currents cancel each other out, leading to:
\(
B=0
\)
Since the magnetic field is zero in all regions (inside the inner conductor, between the conductors, and outside the outer conductor), we conclude that the magnetic field is indeed zero in the entire coaxial cable setup.

Hint

(b, c) As the current is flowing through a conductor so it it is distributed only on the surface of the conductor not in the volume of the cylindrical conductor. It is equivalent to charge distribution on a cylindrical sheet for which electric field inside a conducting cylindrical sheet is always zero.
Magnetic fields at any point inside the conducting cylinder is proportional to the distance from the axis of the cylinder.
At the axis, \(\mathrm{r}=0\). This implies that field will be zero at the axis.

Hint

(b)

\(
\begin{aligned}
&\text { We have } B=\frac{\mu_0}{4 \pi} \cdot \frac{2 I}{r}\\
&\therefore B \propto \frac{1}{r}
\end{aligned}
\)

Hint

(c) Magnetic field at a distance \(r\) near a long straight current carrying wire is given by
\(
\begin{aligned}
B & =\frac{\mu_0}{4 \pi} \cdot \frac{2 i}{r} \Rightarrow B \propto \frac{1}{r} \\
\frac{B_1}{B_2} & =\frac{r_2}{r_1} \text { or } \frac{B}{B_2}=\frac{r / 2}{r} \\
B_2 & =2 B
\end{aligned}
\)

Hint

(c) The correct formula for the magnetic field due to a small current element according to Biot-Savart’s law is \(\frac{\mu_0}{4 \pi} \frac{i \Delta l \sin \theta}{r^2}\).

Hint

(d) To determine the direction of the magnetic field around a current-carrying wire, use the right-hand rule. Place your thumb in the direction of the current (south in this case). Your fingers curled around your thumb represent the direction of the magnetic field. Above the power line, with the thumb pointing south, your fingers curl to the west.

Using Fleming’s right-hand rule, the direction of magnetic field is towards west.

Hint

(a) If a wire carries current along positive \(x\)-axis, then the magnetic field due to it along the line \(y=x\) in first quadrant is upwards. If a wire carries current along positive \(y\)-axis, then the magnetic field due to it along the line \(y=x\) in first quadrant is downwards.
Along the line \(y=x\), above two values of magnetic field are equal in magnitude but opposite in direction and thus, their resultant is zero.
Explanation:
According to right hand screw rule,
Along the line \(x=y\), magnetic field due to two wires will be equal and opposite.

Hint

(b) Magnetic field at the centre of a current carrying circular coil of \(N\) turns is given by
\(
B=\frac{\mu_0 N i}{2 r} \Rightarrow B \propto \frac{1}{r}
\)

Hint

(b) Current produced in wire, \(i=\frac{q}{t}=\frac{100 \times e}{1}\)
\(
\therefore \quad i=100 e
\)
Magnetic field at the centre of circular path,
\(
\begin{aligned}
& =\frac{\mu_0}{4 \pi} \cdot \frac{2 \pi i}{r}=\frac{\mu_0}{4 \pi} \cdot \frac{2 \pi \times 100 e}{r}=\frac{\mu_0 \times 200 e}{4 r} \\
& =\frac{\mu_0 \times 200 \times 1.6 \times 10^{-19}}{4 \times 0.8}=10^{-17} \mu_0 \mathrm{~T}
\end{aligned}
\)

Hint

(c) The direction of magnetic field produced due to both semicircular parts will be perpendicular to the paper and inwards.
Magnetic induction at the centre \(O\),
\(
B=B_1+B_2=\frac{\mu_0 i}{4 r_1}+\frac{\mu_0 i}{4 r_2}=\frac{\mu_0 i}{4}\left(\frac{r_1+r_2}{r_1 r_2}\right)
\)

Hint

(c) Magnetic field at the centre of circular coil of \(N\) turns is given by
\(
B=\frac{\mu_0 N i}{2 r}=\frac{4 \pi \times 10^{-7} \times 100 \times 0.1}{2 \times 5 \times 10^{-2}}=4 \pi \times 10^{-5} \mathrm{~T}
\)

Hint

(d) The given situation is shown below

At centre \(O\), magnetic field due to inner circular coil,
\(
B_1=\frac{\mu_0 N i_1}{2 r_1}=\frac{\mu_0 10 \times(0.2)}{2 \times\left(2 \times 10^{-2}\right)}=50 \mu_0 \odot
\)
Similarly, at centre \(O\), magnetic field due to outer circular coil,
\(
B_2=\frac{\mu_0 N i_2}{2 r_2}=\frac{\mu_0 10 \times(0.3)}{2\left(4 \times 10^{-2}\right)}=\frac{75 \mu_0}{2} \otimes
\)
Net magnetic field,
\(
\begin{aligned}
B_O & =B_1-B_2=50 \mu_0-\frac{75 \mu_0}{2} \\
& =\frac{100 \mu_0-75 \mu_0}{2}=\frac{25}{2} \mu_0 \mathrm{~Wb} / \mathrm{m}^2 \odot
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (c) The magnetic field, } B=\frac{\mu_0 N i}{2 R} \text { or } B \propto \frac{N}{R}\\
&\begin{aligned}
& \frac{B_2}{B_1}=\frac{N_2}{N_1} \times \frac{R_1}{R_2}=\frac{2 N_1}{N_1} \times \frac{R_1}{R_1 / 2}=4 \Rightarrow B_2=4 B_1=4 B \\
& B^{\prime}=4 B
\end{aligned}
\end{aligned}
\)

Hint

(b) Magnetic field at the centre of arc,
\(
B=\frac{\mu_0}{4 \pi} \frac{\theta i}{r}=\frac{\mu_0}{4 \pi} \times \frac{\pi}{2} \times \frac{i}{R}=\frac{\mu_0 i}{8 R}
\)

Hint

(d) Magnetic field induction at centre \(O\) is given by

\(
\begin{aligned}
B & =\frac{3}{4}(\text { Magnetic field due to whole circle }) \\
& =\frac{3}{4}\left(\frac{\mu_0 i}{2 R}\right)=\frac{3 \mu_0 i}{8 R}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (a) The magnetic field at the centre } O \text {, }\\
&\begin{aligned}
B_O & =B_1+B_2 \\
& =\frac{\mu_0 I}{2 R}\left[\frac{2 \pi-2 \theta}{2 \pi}\right]+\frac{\mu_0 I}{4 \pi r}[\sin \theta+\sin \theta] \\
& =\frac{\mu_0 I}{2 R}\left[\frac{\pi-\theta}{\pi}\right]+\frac{\mu_0 I}{4 \pi(R \cos \theta)}(2 \sin \theta) \\
& =\frac{\mu_0 I}{2 \pi R}[\pi-\theta+\tan \theta]
\end{aligned}
\end{aligned}
\)

Hint

(c) Magnetic field due to a ring having \(n\) turns at a distance \(x\) on its axis is given by
\(
\begin{array}{ll}
& B=\frac{\mu_0}{2} \cdot \frac{n i r^2}{\left(x^2+r^2\right)^{3 / 2}} \\
\therefore & B \propto \frac{n r^2}{\left(x^2+r^2\right)^{3 / 2}}
\end{array}
\)

Hint

(d) Magnetic field at the centre of a current carrying coil having current \(I\) and radius \(a\) is given as
\(
B_1=\frac{\mu_0 I}{2 a} \dots(i)
\)
Magnetic field on the axis of circular current carrying coil of radius ‘ \(a\) ‘ and current \(I\) at a distance \(x\) from centre is given as
\(
B_2=\frac{\mu_0 I a^2}{2\left(x^2+a^2\right)^{3 / 2}}
\)
\(
\begin{aligned}
&\text { Here, }\\
&x=a\\
&\therefore \quad \begin{aligned}
B_2 & =\frac{\mu_0 I a^2}{2\left(a^2+a^2\right)^{3 / 2}} \\
& =\frac{\mu_0 I a^2}{2^{5 / 2} \cdot a^3} \\
& =\frac{\mu_0 I}{2^{5 / 2} \cdot a} \dots(ii)
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
&\text { From Eqs. (i) and (ii), we get }\\
&\begin{aligned}
\therefore \quad \frac{B_1}{B_2} & =\frac{\mu_0 I / 2 a}{\mu_0 I / 2^{5 / 2} a}=\frac{2^{5 / 2}}{2} \\
& =2^{3 / 2}=2 \cdot 2^{1 / 2}=2 \sqrt{2}
\end{aligned}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (a) According to the question, }\\
&\begin{aligned}
& & \frac{\mu_0 I R^2}{2\left(x^2+R^2\right)^{3 / 2}} & =\frac{1}{8}\left(\frac{\mu_0 I}{2 R}\right) \\
& \therefore & \frac{R^2}{\left(x^2+R^2\right)^{3 / 2}} & =\frac{1}{8 R} \\
& \text { or } & \left(x^2+R^2\right)^{3 / 2} & =8 R^3 \\
& \text { or } & x^2+R^2 & =4 R^2 \\
& \therefore & x & =\sqrt{3} R
\end{aligned}
\end{aligned}
\)

Hint

(d) Inside the rod, \(B \propto r\) and outside it, \(B \propto \frac{1}{r}\).

Hint

(d) Ampere’s law states that the line integral of the magnetic field \(B\) around a closed loop is equal to the permeability of free space \(\mu_0\) times the total current I passing through the loop. Mathematically, it is expressed as:
\(
\oint B \cdot d l=\mu_0 I_{e n c}
\)
Option a: A circular loop is not suitable because it does not align well with the uniform magnetic field inside the solenoid.
Option b: A rectangular loop perpendicular to the axis of the solenoid results in \(B \cdot d l=0\) since the angle \(\theta\) between \(B\) and \(d l\) is 90 degrees, leading to no contribution to the integral.
Option c: A rectangular loop totally within the solenoid would not allow us to measure the current effectively since the magnetic field lines are uniform and parallel to the axis of the solenoid.
Option d: A rectangular loop that is partially inside and partially outside the solenoid allows us to utilize Ampere’s law effectively. The segment of the loop inside the solenoid will experience the magnetic field, while the segments outside will not contribute to the integral.

Hint

(b) Mean radius of toroid,
\(
r=\frac{25+26}{2}=25.5 \mathrm{~cm}=0.255 \mathrm{~m}
\)
Total number of turns, \(N=3500\)
Current, \(I=11 \mathrm{~A}\)
Number of turns per unit length, \(n=\frac{N}{2 \pi r}=\frac{3500}{2 \pi \times 0.255}\)
Magnetic field inside the core of the toroid,
\(
\begin{aligned}
B=\mu_0 n I= & 4 \pi \times 10^{-7} \times \frac{3500}{2 \pi \times 0.255} \times 11 \\
& =3.0 \times 10^{-2} \mathrm{~T}
\end{aligned}
\)

Hint

(c) For stationary electron, \(v=0\)
\(\therefore \quad F=B q v \sin \theta=B q \times 0 \times \sin \theta=0\), hence on a static charge, there is no effect of magnetic field, so electron remains stationary.

Hint

\(
\text { (c) Path is circular when } \mathbf{v} \perp \mathbf{B} \text { or } \mathbf{v} \cdot \mathbf{B}=0 \text {. }
\)

Hint

(a) Radius of circular path, \(r=\frac{m v}{q B}\)
For electron, \(r=\frac{m v}{e B}\)
\(
\therefore \quad \frac{e}{m}=\frac{v}{B r}
\)

Hint

(c) Radius of circular path of charged particle,
\(
r=\frac{m v}{q B}=\frac{p}{q B}
\)
Since, electron and proton both have same momentum, therefore, the circular path of both will have the same radius.

Hint

(b) As, \(\mathbf{F}_e=\mathbf{F}_m, q \mathbf{E}=q(\mathbf{v} \times \mathbf{B})\)
\(
\therefore \quad \mathbf{E}=\mathbf{v} \times \mathbf{B} \Rightarrow|\mathbf{E}|=v|\mathbf{B}|
\)
Therefore, the two fields ( \(\mathbf{E}\) and \(\mathbf{B}\) ) are perpendicular.

Hint

(a) Work done by magnetic force is zero, hence according to work-energy theorem,
Change in kinetic energy \(=\) Work done=0
So, kinetic energy remains constant.

Hint

(a) Radius of circular path of proton,
\(
\begin{aligned}
r & =\frac{\sqrt{2 K m}}{B q} \text { or } K \propto \frac{q^2}{m} \text { (for same radius) } \\
\Rightarrow \quad K_\alpha & =K\left(\frac{q^{\prime}}{q}\right)\left(\frac{m}{m^{\prime}}\right)=8(2)\left(\frac{1}{4}\right)=4 \mathrm{eV}
\end{aligned}
\)

Hint

(b) When charged particle moving in circular path enters into a region of magnetic field, then time period of charged particle is given by
\(
T=\frac{2 \pi m}{q B}
\)
\(
\therefore \quad T \propto m \quad \text { ( } q \text { and } B \text { are same) }
\)
\(\because \quad m_p>m_e\)
\(\therefore\) Time period of proton \(>\) Time period of electron.

Hint

(d) Time period of charged particle,
\(
T=\frac{2 \pi m}{B q}
\)
Hence, \(T\) is independent of \(r\).

Hint

(c) Radius of particle, \(r=\frac{\sqrt{2 K m}}{B q}\)
\(\therefore\) Area bounded by the path, \(A=\pi r^2=\frac{2 \pi K m}{B^2 q^2}\) or \(A \propto K\)

Hint

(b) Velocity is in \(X Y\)-plane and magnetic field along \(Z\)-axis. Therefore, path of the electron in magnetic field will be a circle. Magnetic force cannot change the speed of a particle but direction of its motion continuously changes. Hence, path of electron will change.

Hint

(a) Radius, \(r=\frac{\sqrt{2 K m}}{B q}\) or \(r \propto \frac{\sqrt{m}}{q}\)
\(
\left(\frac{\sqrt{m}}{q}\right)_\alpha:\left(\frac{\sqrt{m}}{q}\right)_p:\left(\frac{\sqrt{m}}{q}\right)_d=\frac{\sqrt{4}}{2}: \frac{\sqrt{1}}{1}: \frac{\sqrt{2}}{1}=1: 1: \sqrt{2}
\)
Hence, \(r_\alpha=r_p<r_{d}\).

Hint

\(
\begin{aligned}
&\text { (b) Radius of helical path taken by proton beam, }\\
&\begin{aligned}
r=\frac{M v \sin 60^{\circ}}{B q} & =\frac{\left(1.67 \times 10^{-27}\right) \times\left(4 \times 10^5\right) \sin 60^{\circ}}{0.3 \times 1.6 \times 10^{-19}} \\
& =0.012 \mathrm{~m}
\end{aligned}
\end{aligned}
\)

Hint

(a) For charged particle on circular path,
\(
\begin{aligned}
& & \frac{m v^2}{r} & =q v B \\
& \therefore & r & =\frac{m v}{q B}=\frac{\sqrt{2 m E}}{q B}
\end{aligned}
\)
For proton, \(R_p=\frac{\sqrt{2 m_p E}}{q B} \dots(i)\)
For deuteron, \(R_d=\frac{\sqrt{2 m_d E}}{q B} \dots(ii)\)
Dividing Eq. (ii) by Eq. (i), we get
\(
\begin{aligned}
& \frac{R_d}{R_p} & =\sqrt{\frac{m_d}{m_p}}=\sqrt{2} \left[\because m_d=2 m_p\right] \\
\therefore & R_d & =\sqrt{2} R_p
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (a) Radius of the circular path, } r=\frac{m v}{q B}\\
&\Rightarrow \quad \frac{r_1}{r_2}=\frac{m_1 v_1}{m_2 v_2} \times \frac{q_2}{q_1}=\frac{1 \times 2}{1 \times 3} \times \frac{2}{1}=\frac{4}{3}
\end{aligned}
\)

Hint

(a) Frequency of revolution, \(f=\frac{B q}{2 \pi m}\)
\(
f \propto \frac{q}{m}
\)
As \(q\) is same for all given particles, \(f \propto \frac{1}{m}\).
Mass of \(\mathrm{Li}^{+}\)is maximum, therefore its frequency will be minimum.

Hint

(b) Applying Fleming’s left-hand rule, if magnetic field is perpendicular to paper inwards and current in the loop is clockwise, the magnetic force \(F\) on each element of loop is radially outwards and loop has a tendency to expand outwards. Also, when a current carrying loop is placed in uniform magnetic field, then net force on it is zero and loop cannot have translation motion.

Hint

(d) The parallel wires carrying currents in the same direction attract each other because magnetic forces on the two wires act towards each other.

Hint

(b)

\(
\begin{aligned}
&\text { The magnitude of force acting on length } l \text { of each conductor due to other is }\\
&\frac{F}{l}=\frac{\mu_0}{2 \pi} \frac{i_1 i_2}{r}
\end{aligned}
\)
Given \(i_1=i, i_2=10 i\)
\(
\therefore \frac{F}{l}=\frac{\mu_0}{2 \pi} \frac{10 i^2}{r}
\)
From Fleming’s left hand rule, this is force of attraction.

Hint

\(
\begin{aligned}
&\text { (c) Force per unit length, }\\
&\begin{aligned}
& \frac{F}{l}=\frac{\mu_0}{4 \pi} \cdot \frac{2 i_1 i_2}{d} \\
& \frac{F}{l}=\frac{\mu_0}{4 \pi} \cdot \frac{2 i^2}{d} \left(\because i=i_1=i_2\right) \\
& \frac{F}{l}=\frac{\mu_0}{2 \pi} \cdot \frac{i^2}{d} \text { (attractive) }
\end{aligned}
\end{aligned}
\)

Hint

(d) Force between two conductors, \(F \propto \frac{I_1 I_2}{d}\)
When \(I_1\) is changed to \(2 I_1\) and \(d\) is changed to \(3 d\).
\(
\therefore \quad F^{\prime} \propto \frac{\left(2 I_1\right)\left(I_2\right)}{3 d} \propto \frac{2}{3} F
\)
As, direction of current is reversed, so \(F^{\prime}=-2 F / 3\).

Hint

(c) Force per unit length, \(\frac{F}{l}=\frac{\mu_0}{2 \pi} \frac{i_A i_B}{r}\)
Given,
\(
\begin{aligned}
& \quad i_A=2 i_B \\
& (0.004)=\frac{2 \times 10^{-7}}{0.01}\left(2 i_B^2\right) \\
& i_B=10 \mathrm{~A}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { (c) Torque, } \tau_{\max }=N i A B=1 \times i \times \pi r^2 \times B \\
& \therefore \quad\left(\because 2 \pi r=L \Rightarrow r=\frac{L}{2 \pi}\right) \\
& \therefore \quad \tau_{\max }=\pi i\left(\frac{L}{2 \pi}\right)^2 B=\frac{L^2 i B}{4 \pi}
\end{aligned}
\)

Hint

(d) Torque acting on the coil is given by
\(
\tau=N i B A \sin \theta
\)
As, magnetic field is normal to the plane of coil.
\(
\begin{array}{ll}
\text { So, } & \theta=0^{\circ} \\
\therefore & \tau=0
\end{array}
\)

Hint

(d) The pole pieces of the magnet used in a pivoted coil galvanometer are cylindrical surfaces of a horse-shoe magnet.

Hint

\(
\begin{aligned}
&\text { (b) The electrical current in moving coil galvanometer }\\
&\begin{array}{ll}
& i=\frac{k \theta}{N A B} \\
\Rightarrow \quad & i \propto \theta
\end{array}
\end{aligned}
\)

Hint

(b) We have sensitivity of a moving coil galvanometer
\(
S=\frac{N A B}{k}
\)
where, \(k\) is the torsional constant of its suspension.
In order to increase the sensitivity of a moving coil galvanometer, one should decrease the torsional constant of its suspension.

Hint

\(
\begin{aligned}
&\text { (a) Sensitivity, } S=\frac{\theta}{i}\\
&\frac{S_A}{S_B}=\frac{i_B}{i_A} \Rightarrow \frac{S_A}{S_B}=\frac{5}{3} \Rightarrow S_A>S_B
\end{aligned}
\)