NCERT Exercise Q & A

NCERT Exercise Problems

Q4.1: A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A . What is the magnitude of the magnetic field \(\mathbf{B}\) at the centre of the coil?

Answer: Number of turns on the circular coil, \(n=100\)
Radius of each turn, \(r=8.0 \mathrm{~cm}=0.08 \mathrm{~m}\)
Current flowing in the coil, \(I=0.4 \mathrm{~A}\)
Magnitude of the magnetic field at the centre of the coil is given by the relation,
\(
|\mathbf{B}|=\frac{\mu_0}{4 \pi} \frac{2 \pi n I}{r}
\)
Where,
\(
\begin{aligned}
& \mu_0=\text { Permeability of free space } \\
& =4 \pi \times 10^{-7} \mathrm{~T} \mathrm{~m} \mathrm{~A}^{-1} \\
& \begin{aligned}
|\mathbf{B}| & =\frac{4 \pi \times 10^{-7}}{4 \pi} \times \frac{2 \pi \times 100 \times 0.4}{0.08} \\
& =3.14 \times 10^{-4} \mathrm{~T}
\end{aligned}
\end{aligned}
\)
Hence, the magnitude of the magnetic field is \(3.14 \times 10^{-4} \mathrm{~T}\).

Q4.2: A long straight wire carries a current of 35 A . What is the magnitude of the field \(\mathbf{B}\) at a point 20 cm from the wire?

Answer: Current in the wire, \(I=35 \mathrm{~A}\)
Distance of a point from the wire, \(r=20 \mathrm{~cm}=0.2 \mathrm{~m}\)
Magnitude of the magnetic field at this point is given as:
\(
B=\frac{\mu_0}{4 \pi} \frac{2 I}{r}
\)
Where,
\(
\begin{aligned}
\mu_0 & =\text { Permeability of free space }=4 \pi \times 10^{-7} \mathrm{~T} \mathrm{~m} \mathrm{~A}^{-1} \\
B & =\frac{4 \pi \times 10^{-7} \times 2 \times 35}{4 \pi \times 0.2} \\
& =3.5 \times 10^{-5} \mathrm{~T}
\end{aligned}
\)
Hence, the magnitude of the magnetic field at a point 20 cm from the wire is \(3.5 \times 10^{-5}\) T.

Q4.3: A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of \(\mathbf{B}\) at a point 2.5 m east of the wire.

Answer: Current in the wire, \(I=50 \mathrm{~A}\)
A point is 2.5 m away from the East of the wire.
\(\therefore\) Magnitude of the distance of the point from the wire, \(r=2.5 \mathrm{~m}\).
Magnitude of the magnetic field at that point is given by the relation, \(B=\frac{\mu_0 2 I}{4 \pi r}\)
Where,
\(
\begin{aligned}
\mu_0 & =\text { Permeability of free space }=4 \pi \times 10^{-7} \mathrm{~T} \mathrm{~m} \mathrm{~A}^{-1} \\
B & =\frac{4 \pi \times 10^{-7} \times 2 \times 50}{4 \pi \times 2.5} \\
& =4 \times 10^{-6} \mathrm{~T}
\end{aligned}
\)
The point is located normal to the wire length at a distance of 2.5 m . The direction of the current in the wire is vertically downward. Hence, according to the Maxwell’s right hand thumb rule, the direction of the magnetic field at the given point is vertically upward.

Q4.4: A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?

Answer: Current in the power line, \(I=90 \mathrm{~A}\)
Point is located below the power line at distance, \(r=1.5 \mathrm{~m}\)
Hence, magnetic field at that point is given by the relation,
\(
B=\frac{\mu_0 2 I}{4 \pi r}
\)
Where,
\(
\begin{aligned}
& \mu_0=\text { Permeability of free space }=4 \pi \times 10^{-7} \mathrm{~T} \mathrm{~m} \mathrm{~A}^{-1} \\
& B=\frac{4 \pi \times 10^{-7} \times 2 \times 90}{4 \pi \times 1.5}=1.2 \times 10^{-5} \mathrm{~T}
\end{aligned}
\)
The current is flowing from East to West. The point is below the power line. Hence, according to Maxwell’s right hand thumb rule, the direction of the magnetic field is towards the South.

Q4.5: What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of \(30^{\circ}\) with the direction of a uniform magnetic field of 0.15 T ?

Answer: Current in the wire, \(I=8 \mathrm{~A}\)
Magnitude of the uniform magnetic field, \(B=0.15 \mathrm{~T}\)
Angle between the wire and magnetic field, \(\theta=30^{\circ}\).
Magnetic force per unit length on the wire is given as:
\(
\begin{aligned}
& f=B I \sin \theta \\
& =0.15 \times 8 \times 1 \times \sin 30^{\circ} \\
& =0.6 \mathrm{~N} \mathrm{~m}^{-1}
\end{aligned}
\)
Hence, the magnetic force per unit length on the wire is \(0.6 \mathrm{~N} \mathrm{~m}^{-1}\).

Q4.6: A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T . What is the magnetic force on the wire?

Answer: Length of the wire, \(l=3 \mathrm{~cm}=0.03 \mathrm{~m}\)
Current flowing in the wire, \(I=10 \mathrm{~A}\)
Magnetic field, \(B=0.27 \mathrm{~T}\)
Angle between the current and magnetic field, \(\theta=90^{\circ}\)
Magnetic force exerted on the wire is given as:
\(
\begin{aligned}
& F=B I l \sin \theta \\
& =0.27 \times 10 \times 0.03 \sin 90^{\circ} \\
& =8.1 \times 10^{-2} \mathrm{~N}
\end{aligned}
\)
Hence, the magnetic force on the wire is \(8.1 \times 10^{-2} \mathrm{~N}\). The direction of the force can be obtained from Fleming’s left hand rule.

Q4.7: Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm . Estimate the force on a 10 cm section of wire A.

Answer: Current flowing in wire \(\mathrm{A}, I_{\mathrm{A}}=8.0 \mathrm{~A}\)
Current flowing in wire \(\mathrm{B}, I_{\mathrm{B}}=5.0 \mathrm{~A}\)
Distance between the two wires, \(r=4.0 \mathrm{~cm}=0.04 \mathrm{~m}\)
Length of a section of wire \(A, I=10 \mathrm{~cm}=0.1 \mathrm{~m}\)
Force exerted on length / due to the magnetic field is given as:
\(
B=\frac{\mu_0 2 I_{\mathrm{A}} I_{\mathrm{B}} l}{4 \pi r}
\)
Where,
\(
\mu_0=\text { Permeability of free space }=4 \pi \times 10^{-7} \mathrm{~T} \mathrm{~m} \mathrm{~A}
\)
\(
\begin{aligned}
B & =\frac{4 \pi \times 10^{-7} \times 2 \times 8 \times 5 \times 0.1}{4 \pi \times 0.04} \\
& =2 \times 10^{-5} \mathrm{~N}
\end{aligned}
\)
The magnitude of force is \(2 \times 10^{-5} \mathrm{~N}\). This is an attractive force normal to A towards B because the direction of the currents in the wires is the same.

Q4.8: A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm . If the current carried is 8.0 A , estimate the magnitude of \(\mathbf{B}\) inside the solenoid near its centre.

Answer: Length of the solenoid, \(l=80 \mathrm{~cm}=0.8 \mathrm{~m}\)
There are five layers of windings of 400 turns each on the solenoid.
\(\therefore\) Total number of turns on the solenoid, \(N=5 \times 400=2000\)
Diameter of the solenoid, \(D=1.8 \mathrm{~cm}=0.018 \mathrm{~m}\)
Current carried by the solenoid, \(I=8.0 \mathrm{~A}\)
Magnitude of the magnetic field inside the solenoid near its centre is given by the relation,
\(
B=\frac{\mu_0 N I}{l}
\)
Where,
\(
\begin{aligned}
\mu_0 & =\text { Permeability of free space }=4 \pi \times 10^{-7} \mathrm{~T} \mathrm{~m} \mathrm{~A}^{-1} \\
B & =\frac{4 \pi \times 10^{-7} \times 2000 \times 8}{0.8} \\
& =8 \pi \times 10^{-3}=2.512 \times 10^{-2} \mathrm{~T}
\end{aligned}
\)
Hence, the magnitude of the magnetic field inside the solenoid near its centre is \(2.512 \times\) \(10^{-2} \mathrm{~T}\)

Q4.9: A square coil of side 10 cm consists of 20 turns and carries a current of 12 A . The coil is suspended vertically and the normal to the plane of the coil makes an angle of \(30^{\circ}\) with the direction of a uniform horizontal magnetic field of magnitude 0.80 T . What is the magnitude of torque experienced by the coil?

Answer: Length of a side of the square coil, \(l=10 \mathrm{~cm}=0.1 \mathrm{~m}\)
Current flowing in the coil, \(I=12 \mathrm{~A}\)
Number of turns on the coil, \(n=20\)
Angle made by the plane of the coil with magnetic field, \(\theta=30^{\circ}\)
Strength of magnetic field, \(B=0.80 \mathrm{~T}\)
Magnitude of the magnetic torque experienced by the coil in the magnetic field is given by the relation,
\(
\mathrm{T}=n B I A \sin \theta
\)
Where,
\(A=\) Area of the square coil
\(
\Rightarrow l \times l=0.1 \times 0.1=0.01 \mathrm{~m}^2
\)
\(
\therefore \mathrm{T}=20 \times 0.8 \times 12 \times 0.01 \times \sin 30^{\circ}
\)
\(
=0.96 \mathrm{~N} \mathrm{~m}
\)
Hence, the magnitude of the torque experienced by the coil is 0.96 N m .

Q4.10: Two moving coil meters, \(M_1\) and \(M_2\) have the following particulars:
\(
\begin{aligned}
& R_1=10 \Omega, N_1=30 \\
& A_1=3.6 \times 10^{-3} \mathrm{~m}^2, B_1=0.25 \mathrm{~T} \\
& R_2=14 \Omega, N_2=42 \\
& A_2=1.8 \times 10^{-3} \mathrm{~m}^2, B_2=0.50 \mathrm{~T}
\end{aligned}
\)
(The spring constants are identical for the two meters).
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of \(M_2\) and \(M_1\).

Answer: For moving coil meter \(M_1\) :
Resistance, \(R_1=10 \Omega\)
Number of turns, \(N_1=30\)
Area of cross-section, \(A_1=3.6 \times 10^{-3} \mathrm{~m}^2\)
Magnetic field strength, \(B_1=0.25 \mathrm{~T}\)
Spring constant \(K_1=K\)
For moving coil meter \(M_2\) :
Resistance, \(R_2=14 \Omega\)
Number of turns, \(N_2=42\)
Area of cross-section, \(A_2=1.8 \times 10^{-3} \mathrm{~m}^2\)
Magnetic field strength, \(B_2=0.50 \mathrm{~T}\)
Spring constant, \(K_2=K\)
(a) Current sensitivity of \(M_1\) is given as:
\(
I_{\mathrm{s1}}=\frac{N_1 B_1 A_1}{K_1}
\)
And, current sensitivity of \(\mathrm{M}_2\) is given as:
\(
\begin{aligned}
& I_{\mathrm{s} 2}=\frac{N_2 B_2 A_2}{K_2} \\
& \therefore \text { Ratio } \frac{I_{\mathrm{s} 2}}{I_{\mathrm{s} 1}}=\frac{N_2 B_2 A_2 K_1}{K_2 N_1 B_1 A_1} \\
& =\frac{42 \times 0.5 \times 1.8 \times 10^{-3} \times K}{K \times 30 \times 0.25 \times 3.6 \times 10^{-3}}=1.4
\end{aligned}
\)
Hence, the ratio of current sensitivity of \(M_2\) to \(M_1\) is 1.4 .
(b) Voltage sensitivity for \(M_2\) is given as:
\(
V_{\mathrm{s} 2}=\frac{N_2 B_2 A_2}{K_2 R_2}
\)
And, voltage sensitivity for \(M_1\) is given as:
\(
\begin{aligned}
& V_{\mathrm{sl}}=\frac{N_1 B_1 A_1}{K_1} \\
& \therefore \frac{V_{\mathrm{s} 2}}{V_{\mathrm{s} 1}}=\frac{N_2 B_2 A_2 K_1 R_1}{K_2 R_2 N_1 B_1 A_1} \\
& =\frac{42 \times 0.5 \times 1.8 \times 10^{-3} \times 10 \times K}{K \times 14 \times 30 \times 0.25 \times 3.6 \times 10^{-3}}=1
\end{aligned}
\)
Hence, the ratio of voltage sensitivity of \(M_2\) to \(M_1\) is 1 .

Q4.11: In a chamber, a uniform magnetic field of \(6.5 \mathrm{G}\left(1 \mathrm{G}=10^{-4} \mathrm{~T}\right)\) is maintained. An electron is shot into the field with a speed of \(4.8 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}\) normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. \(\left(e=1.5 \times 10^{-19} \mathrm{C}, m_e=9.1 \times 10^{-31} \mathrm{~kg}\right)\)

Answer: Magnetic field strength, \(B=6.5 \mathrm{G}=6.5 \times 10^{-4} \mathrm{~T}\)
Speed of the electron, \(v=4.8 \times 10^6 \mathrm{~m} / \mathrm{s}\)
Charge on the electron, \(e=1.6 \times 10^{-19} \mathrm{C}\)
Mass of the electron, \(m_e=9.1 \times 10^{-31} \mathrm{~kg}\)
Angle between the shot electron and magnetic field, \(\theta=90^{\circ}\)
Magnetic force exerted on the electron in the magnetic field is given as:
\(
F=e v B \sin \theta
\)
This force provides centripetal force to the moving electron. Hence, the electron starts moving in a circular path of radius \(r\).
Hence, centripetal force exerted on the electron,
\(
F_{\mathrm{c}}=\frac{m v^2}{r}
\)
In equilibrium, the centripetal force exerted on the electron is equal to the magnetic force i.e.,
\(
\begin{aligned}
F_{\mathrm{c}} & =F \\
\frac{m v^2}{r} & =e v B \sin \theta \\
r & =\frac{m v}{B e \sin \theta} \\
& =\frac{9.1 \times 10^{-31} \times 4.8 \times 10^6}{6.5 \times 10^{-4} \times 1.6 \times 10^{-19} \times \sin 90^{\circ}} \\
& =4.2 \times 10^{-2} \mathrm{~m}=4.2 \mathrm{~cm}
\end{aligned}
\)
Hence, the radius of the circular orbit of the electron is 4.2 cm .

Q4.12: In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.

Answer: Magnetic field strength, \(B=6.5 \times 10^{-4} \mathrm{~T}\)
Charge of the electron, \(e=1.6 \times 10^{-19} \mathrm{C}\)
Mass of the electron, \(m_e=9.1 \times 10^{-31} \mathrm{~kg}\)
Velocity of the electron, \(v=4.8 \times 10^6 \mathrm{~m} / \mathrm{s}\)
Radius of the orbit, \(r=4.2 \mathrm{~cm}=0.042 \mathrm{~m}\)
Frequency of revolution of the electron \(=\nu\)
Angular frequency of the electron \(=\omega=2 \pi \nu\)
Velocity of the electron is related to the angular frequency as:
\(
v=r \omega
\)
In the circular orbit, the magnetic force on the electron is balanced by the centripetal force. Hence, we can write:
\(
\begin{aligned}
& e v B=\frac{m v^2}{r} \\
& e B=\frac{m}{r}(r \omega)=\frac{m}{r}(r 2 \pi \nu) \\
& \nu=\frac{B e}{2 \pi m}
\end{aligned}
\)
This expression for frequency is independent of the speed of the electron.
On substituting the known values in this expression, we get the frequency as:
\(
\begin{aligned}
\nu & =\frac{6.5 \times 10^{-4} \times 1.6 \times 10^{-19}}{2 \times 3.14 \times 9.1 \times 10^{-31}} \\
& =18.2 \times 10^6 \mathrm{~Hz} \\
& \approx 18 \mathrm{MHz}
\end{aligned}
\)
Hence, the frequency of the electron is around 18 MHz and is independent of the speed of the electron.

Q4.13: (a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T . The field lines make an angle of \(60^{\circ}\) with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.
(b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)

Answer: (a) Number of turns on the circular coil, \(n=30\)
Radius of the coil, \(r=8.0 \mathrm{~cm}=0.08 \mathrm{~m}\)
Area of the coil \(=\pi r^2=\pi(0.08)^2=0.0201 \mathrm{~m}^2\)
Current flowing in the coil, \(I=6.0 \mathrm{~A}\)
Magnetic field strength, \(B=1 \mathrm{~T}\)
Angle between the field lines and normal with the coil surface,
\(
\theta=60^{\circ}
\)
The coil experiences a torque in the magnetic field. Hence, it turns. The counter torque applied to prevent the coil from turning is given by the relation,
\(
\begin{aligned}
& \mathrm{T}=n I B A \sin \theta \ldots(i) \\
& =30 \times 6 \times 1 \times 0.0201 \times \sin 60^{\circ} \\
& =3.133 \mathrm{~N} \mathrm{~m}
\end{aligned}
\)
(b) It can be inferred from relation (i) that the magnitude of the applied torque is not dependent on the shape of the coil. It depends on the area of the coil. Hence, the answer would not change if the circular coil in the above case is replaced by a planar coil of some irregular shape that encloses the same area.

Exemplar Problems

VSA

Q4.12: Verify that the cyclotron frequency \(\omega=e B / m\) has the correct dimensions of \([T]^{-1}\). 

Answer: In cyclotron, charge particle describes the circular path where magnetic force acts as centripetal force.
\(
\frac{m v^2}{R}=e v B
\)
On simplifying the terms, we have
\(
\therefore \quad \frac{e B}{m}=\frac{v}{R}=\omega
\)
We know that \(B=\frac{F}{e v}=\frac{\left[M L T^{-2}\right]}{[A T]\left[L T^{-1}\right]}=\left[M A^{-1} T^{-2}\right]\)
And then dimensional formula of angular frequency
\(
\begin{aligned}
\therefore \quad[\omega] & =\left[\frac{e B}{m}\right]=\left[\frac{v}{R}\right] \\
{[\omega] } & =\frac{[A T]\left[M A^{-1} T^{-2}\right]}{[M]}=\left[T^{-1}\right]
\end{aligned}
\)

Q4.13: Show that a force that does no work must be a velocity dependent force.

Answer: To show that a force that does no work must be a velocity dependent force, then we have to assume that work done by force is zero. As shown by the equation below:
\(
d W=\vec{F} \cdot \overrightarrow{d l}=0
\)
We can write, \(\overrightarrow{d l}=\vec{v} d t\), but \(d t \neq 0\)
\(
\begin{aligned}
& \Rightarrow \quad \vec{F} \cdot \vec{v} d t=0 \\
& \Rightarrow \quad \vec{F} \cdot \vec{v}=0
\end{aligned}
\)
So we can say that force F must be velocity dependent, this implies that angle between F and vis \(90^{\circ}\). If the direction of velocity changes, then direction of force will also change.

Q4.14: The magnetic force depends on \(\mathbf{v}\) which depends on the inertial frame of reference. Does then the magnetic force differ from inertial frame to frame? Is it reasonable that the net acceleration has a different value in different frames of reference?

Answer: Magnetic force a charged particle, \(\vec{F}=q(\vec{v} \times \vec{B})\) or \(F=q v B \sin \theta\). Thus the magnetic force is a velocity dependent. Thus the magnetic force is a velocity dependent. It differs from one inertial frame to another. The net acceleration arising due to magnetic force is independent from frame to frame for inertial frames of reference.

Q4.15: Describe the motion of a charged particle in a cyclotron if the frequency of the radio frequency (rf) field were doubled.

Answer: The frequency \(\mathrm{v}_{\mathrm{a}}\) of the applied voltage (radio frequency) is adjusted so that the polarity of the dees is reversed at the same time that it takes the ions to complete one-half of the revolution. The requirement \(\mathrm{v}_{\mathrm{a}}=\mathrm{v}_{\mathrm{c}}\) is called the resonance condition.

When the frequency of the radio frequency (rf) field was doubled, then the resonance condition are violated and the time period of the radio frequency (rf) field was halved. Therefore, the duration in which a particle completes half revolution inside the dees, radio frequency completes the cycle. So, particles will accelerate and decelerate alternatively. So, the radius of path in the dees will remain the same.

Q4.16: Two long wires carrying current \(I_1\) and \(I_2\) are arranged as shown in Fig. 4.1. The one carrying current \(I_1\) is along is the \(x\)-axis. The other carrying current \(I_2\) is along a line parallel to the \(y\)-axis given by \(x=0\) and \(z=d\). Find the force exerted at \(\mathrm{O}_2\) because of the wire along the \(x\)-axis.

Answer:In this problem first we have to find the direction of magnetic field due to one wire at the point on other wire, then the magnetic force on that current carrying wire. In Biot-Savart law, magnetic field B is parallel to; \(\mathrm{dl} \times \mathrm{r}\) and idl have its direction along the direction of flow of current, or we can find the direction of \(B\) with the help of right-hand thumb rule.
We know that force on current \((\mathrm{I})\) carrying conductor placed in magnetic field \(\mathrm{B}\) is \(\mathrm{F}=\mathrm{I}(\mathrm{L} \times \mathrm{B})\) \(=I L B \sin \theta\)
The direction of magnetic field at \(\mathrm{O}_2\) due to the current \(\mathrm{I}_1\) is parallel to Y-axis and in -Y direction.

As wire of current \(I_2\) is parallel to Y-axis, current in \(I_2\) is also along Y-axis. So \(I_2\) and \(B_1\) (magnetic field due to current \(I_1\) ) are also along \(Y\)-axis i.e., angle between \(I_2\) and \(B_1\) is zero. So magnetic force \(\mathrm{F}_2\) on wire of current \(\mathrm{I}_2\) is \(\mathrm{F}_2=\mathrm{B}_1 \mathrm{I}_2 \mathrm{~L}_1 \sin 0^{\circ}=0\).

Short Answer Type (SA)

Q4.17: A current carrying loop consists of 3 identical quarter circles of radius \(R\), lying in the positive quadrants of the \(x-y, y-z\) and \(z-x\) planes with their centres at the origin, joined together. Find the direction and magnitude of \(\mathbf{B}\) at the origin.

Answer: Key concept: From Biot-Savart law we find the relation of magnetic field at centre of the current carrying coil which subtends an angle \(\theta, B=\frac{\mu_0}{4 \pi} \frac{I \theta}{R}\).

Magnetic field at origin due to the quarter circle lying in \(x-y\) plane:
\(
\vec{B}_1=\frac{\mu_0}{4 \pi} \frac{I(\pi / 2)}{R} \hat{k}=\frac{\mu_0}{4} \frac{I}{2 R} \hat{k}
\)
Similarly, magnetic field at origin due to the quarter circle lying in the \(y-z\) plane:
\(
\vec{B}_2=\left(\frac{\mu_0}{4}\right) \frac{I}{2 R} \hat{i}
\)
Similarly, magnetic field at origin due to the quarter circle lying in the \(z-x\) plane:
\(
\vec{B}_3=\left(\frac{\mu_0}{4}\right) \frac{I}{2 R} \hat{j}
\)
Now, vector sum of magnetic field at origin due to each quarter is given by
\(
\vec{B}_{\text {net }}=\vec{B}_1+\vec{B}_2+\vec{B}_3=\frac{1}{4}\left(\frac{\mu_0 I}{2 R}\right)(\hat{i}+\hat{j}+\hat{k})
\)

Q4.18: A charged particle of charge \(e\) and mass \(m\) is moving in an electric field \(\mathbf{E}\) and magnetic field \(\mathbf{B}\). Construct dimensionless quantities and quantities of dimension \([T]^{-1}\).

Answer: When a charged particle is moving in both electric and magnetic field, the physical quantities can no help to construct dimensional quantity.
For a charged particle moving perpendicular to the magnetic field, the magnetic Lorentz forces provides necessary centripetal force for revolution.
\(
\mathrm{F}_{\mathrm{m}}=\mathrm{qvB} \sin 90^{\circ}=\mathrm{qvB} \dots(1)
\)
\(
\text { we know that Centripetal force }=\frac{m v^2}{R} \dots(2)
\)
By eq. (1) and (2) \(\frac{m v^2}{R}=q v B\)
\(\because \mathrm{v}=\omega \mathrm{R}\) and \(\mathrm{q}=\mathrm{e}\)
amgular velocity \(\omega=\frac{\nu}{R}=\frac{e B}{m}\)
Dimensional formula for angular velocity \(\omega\)
\(
\omega=\left[\frac{e B}{m}\right]=\left[\frac{v}{R}\right]=\left[T^{-1}\right]
\)

Q4.19: An electron enters with a velocity \(\mathbf{v}=v_0 \mathbf{i}\) into a cubical region (faces parallel to coordinate planes) in which there are uniform electric and magnetic fields. The orbit of the electron is found to spiral down inside the cube in plane parallel to the \(x-y\) plane. Suggest a configuration of fields \(\mathbf{E}\) and \(\mathbf{B}\) that can lead to it.

Answer: The velocity of electron is \(\mathbf{v}=v_0 \mathbf{i}\), i.e., along \(X\)-axis as magnetic field is perpendicular to velocity so it is in Y-direction.
The moving electron enters into cubical region. The force on electron due to Lorentz force
\(
F=q(E+v \times B)
\)
By putting the values
\(
\mathrm{F}_{\mathrm{m}}=-e\left[v_o \hat{i} \times b \widehat{k}\right]=-e v_0 B \hat{j}
\)
Which revolves around the electron in X-Y plane.
The force due to electric field \(F_m=e \vec{K} \widehat{k}\) accelerates electron along z-axis and force due to magnetic field keeps it in circular motion, which in turn increases the radius of circular path. So the motion becomes helical path.

Which revolves the electron in \(x-y\) plane. The electric force \(\mathrm{F}=\mathrm{eE}_0 \mathrm{j}\) accelerates e along the z -axis which in turn increases the radius of circular path and hence particle traversed on spiral path.

Q4.20: Do magnetic forces obey Newton’s third law. Verify for two current elements \(\boldsymbol{d} \boldsymbol{l}_1=d l \hat{\mathbf{i}}\) located at the origin and \(\boldsymbol{d l _ { 2 }}=d l \hat{\mathbf{j}}\) located at \((0, R, 0)\). Both carry current \(I\).

Answer: Step 1: Define the Current Elements
We have two current elements: \(d \overrightarrow{l_1}=d l \hat{i}\) located at the origin \((0,0,0)\). \(d \overrightarrow{l_2}=dl \hat{\mathbf{j}}\) located at \((0, R, 0)\).
Both carry a current \(I\).
Step 2: Calculate the Magnetic Field due to \(d \overrightarrow{l_1}\) at the Location of \(d \overrightarrow{l_2}\)
Using the Biot-Savart law, the magnetic field \(\vec{B}_1\) at the position of \(d \vec{l}_2(\mathrm{0}, \mathrm{R}, \mathrm{0})\) due to the current element \(d \vec{l}_1\) can be calculated as:
\(
\vec{B}_1=\frac{\mu_0 I}{4 \pi} \frac{d \vec{l}_1 \times \hat{r}}{r^2}
\)
Where \(\hat{r}\) is the unit vector pointing from \(d \vec{l}_1\) to \(d \vec{l}_2\) and \(r\) is the distance between them.
The distance \(r\) is \(R\), and the unit vector \(\hat{r}\) from \((0,0,0)\) to \((0, R, 0)\) is \(\hat{j}\). Thus:
\(
\overrightarrow{B_1}=\frac{\mu_0 I}{2 \pi R} \hat{k}
\)
Step 3: Calculate the Force on \(d \vec{l}_2\) due to \(\vec{B}_1\)
The force \(\vec{F}_2\) on the current element \(d \vec{l}_2\) in the magnetic field \(\vec{B}_1\) is given by:
\(
\vec{F}_2=I d \vec{l}_2 \times \vec{B}_1
\)
Substituting \(d \overrightarrow{l_2}=d l \hat{j}\) and \(\vec{B}_1=\frac{\mu_0 I}{2 \pi R} \hat{k}\) :
\(
\vec{F}_2=I(d l \hat{j}) \times\left(\frac{\mu_0 I}{2 \pi R} \hat{k}\right)
\)
Using the right-hand rule, we find:
\(
\vec{F}_2=\frac{\mu_0 I^2 d l}{2 \pi R} \hat{i}
\)
Step 4: Calculate the Magnetic Field due to \(d \vec{l}_2\) at the Location of \(d \overrightarrow{l_1}\)
Now, we calculate the magnetic field \(\vec{B}_2\) at the origin \((0,0,0)\) due to \(d \vec{l}_2\) :
\(
\vec{B}_2=\frac{\mu_0 I}{4 \pi} \frac{d \vec{l}_2 \times \hat{r}}{r^2}
\)
Here, \(r=R\) and \(\hat{r}\) points from \((0, R, 0)\) to \((0,0,0)\), which is \(-\hat{j}\). Thus:
\(
\vec{B}_2=0
\)
since \(d \vec{l}_2\) is parallel to \(\hat{j}\) and \(\hat{r}\) is also along \(-\hat{\boldsymbol{j}}\).
Step 5: Calculate the Force on \(d \overrightarrow{l_1}\) due to \(\vec{B}_2\)
Since \(\vec{B}_2=0\) :
\(
\vec{F}_1=I d \vec{l}_1 \times \vec{B}_2=0
\)
Step 6: Conclusion
From the calculations:
The force \(\vec{F}_2\) on \(d \vec{l}_2\) is \(\frac{\mu_0 I^2 d l}{2 \pi R} \hat{i}\).
The force \(\vec{F}_1\) on \(d \overrightarrow{l_1}\) is 0 .
Since \(\vec{F}_1\) and \(\vec{F}_2\) are not equal and opposite, we conclude that magnetic forces do not obey Newton’s third law.

Q4.21: A multirange voltmeter can be constructed by using a galvanometer circuit as shown in Fig. 4.2. We want to construct a voltmeter that can measure \(2 \mathrm{~V}, 20 \mathrm{~V}\) and 200 V using a galvanometer of resistance \(10 \Omega\) and that produces maximum deflection for current of 1 mA . Find \(R_1, R_2\) and \(R_3\) that have to be used.

Answer: Key concept: The galvanometer can also be used as a voltmeter to measure the voltage across a given section of the circuit. For this a very high resistance wire is to be connected in series with galvanometer. The relationship is given by \(\mathrm{I}_{\mathrm{g}}(\mathrm{G}+\mathrm{R})=\mathrm{V}\) where \(\mathrm{I}_{\mathrm{g}}\) is the range of galvanometer, \(\mathrm{G}\) is the resistance of galvanometer and \(\mathrm{R}\) is the resistance of wire connected in series with galvanometer.
Applying expression in different situations
For \(i_G\left(G+R_1\right)=2\) for 2 V range
For \(i_G\left(G+R_1+R_2\right)=20\) for 20 V range
And For \(i_G\left(G+R_1+R_2+R_3\right)=200\) for 200 V range
By solving, we get
\(
R_1=1990 \Omega, R_2=18 \mathrm{k} \Omega \text { and } R_3=180 \mathrm{k} \Omega .
\)

Q4.22: A long straight wire carrying current of 25 A rests on a table as shown in Fig. 4.3. Another wire PQ of length 1 m , mass 2.5 g carries the same current but in the opposite direction. The wire PQ is free to slide up and down. To what height will PQ rise?

Answer: Key concept: The force applied on PQ by a long straight wire carrying current of 25 A which rests on a table. And the forces which other are repulsive if two straight wires are placed parallel to each other carrying current in opposite direction. Now if the wire PQ is in equilibrium then that repulsive force onPQ must balance its weight.

The magnetic field produced by a long straight wire carrying current of 25 A rests on a table on small wire.
\(
B=\frac{\mu_0 I}{2 \pi h}
\)
The magnetic force on small conductor is
\(
F=B I l \sin \theta=B I l
\)
Force applied on \(P Q\) balance the weight of small current carrying wire,
\(
\begin{aligned}
& F=m g=\frac{\mu_0 I^2 l}{2 \pi h} \\
& h=\frac{\mu_0 I^2 l}{2 \pi m g}=\frac{4 \pi \times 10^{-7} \times(25)^2 \times 1}{2 \pi \times 2.5 \times 10^{-3} \times 9.8}=51 \times 10^{-4} \mathrm{~m} \\
& h=0.51 \mathrm{~cm}
\end{aligned}
\)

Long Answer Type (LA)

Q4.23: A 100 turn rectangular coil ABCD (in XY plane) is hung from one arm of a balance (Fig. 4.4). A mass 500 g is added to the other arm to balance the weight of the coil. A current 4.9 A passes through the coil and a constant magnetic field of 0.2 T acting inward (in \(x z\) plane) is switched on such that only arm CD of length 1 cm lies in the field. How much additional mass ‘ \(m\) ‘ must be added to regain the balance?

Answer: Key concept: Here we use the concept of magnetic force on straight current carrying conductor placed in the region of external uniform magnetic field. The magnetic force exerted on CD due to external magnetic field must balance its weight. And spring balance to be in equilibrium net torque should also be equal to zero.
At \(t=0\), the external magnetic field is off. Let us consider the separation of each hung from mid-point be \(l\).
\(
\begin{aligned}
& M g l=W_{\text {coil }} l \\
& 0.5 \mathrm{gl}=W_{\text {coil }} l \\
& W_{\text {coil }}=0.5 \times 9.8 \mathrm{~N}
\end{aligned}
\)
By taking moment of force about mid-point, we get the weight of coil.
And let ‘ \(m\) ‘ be the mass which is added to regain the balance.
When the magnetic field is switched on.
\(
\begin{aligned}
& M g l+m g l=W_{\text {coil }} l+\left(I L B \sin 90^{\circ}\right) l \\
& m g l=(I L B) l \\
& m=\frac{B I L}{g}=\frac{0.2 \times 4.9 \times 1 \times 10^{-2}}{9.8}=10^{-3} \mathrm{~kg}=1 \mathrm{~g}
\end{aligned}
\)
Therefore, 1 g of additional mass must be added to regain the balance.

Q4.24: A rectangular conducting loop consists of two wires on two opposite sides of length \(l\) joined together by rods of length \(d\). The wires are each of the same material but with cross-sections differing by a factor of 2. The thicker wire has a resistance \(R\) and the rods are of low resistance, which in turn are connected to a constant voltage source \(V_0\). The loop is placed in uniform a magnetic field \(\mathbf{B}\) at \(45^{\circ}\) to its plane. Find \(\mathbf{\tau}\), the torque exerted by the magnetic field on the loop about an axis through the centres of rods.

Answer: After analyzing the direction of current in both wires, magnetic forces and torques need to be calculated for finding the net torque.

According to the problem, the thicker wire has a resistance \(R\), then the other wire has a resistance \(2 R\) as the wires are of the same material so their resistivity remains same.
Now, the force and hence, torque on first wire is given by
\(
\begin{aligned}
& F_1=i_1 l B \sin 90^{\circ}=\frac{V_0}{2 R} l B \\
& \tau_1=\frac{d}{2 \sqrt{2}} F_1=\frac{V_0 l d B}{2 \sqrt{2} R}
\end{aligned}
\)
Similarly, the force hence torque on other wire is given by
\(
\begin{aligned}
F_2 & =i_2 l B \sin 90^{\circ}=\frac{V_0}{2 R} l B \\
\tau_2 & =\frac{d}{2 \sqrt{2}} F_2=\frac{V_0 l d B}{4 \sqrt{2} R}
\end{aligned}
\)
So, net torque, \(\tau=\tau_1-\tau_2\)
\(
\tau=\frac{1}{4 \sqrt{2}} \frac{V_0 A B}{R}
\)
where \(A\) is the area of rectangular coil.

Q4.25: An electron and a positron are released from \((0,0,0)\) and \((0,0,1.5 R)\) respectively, in a uniform magnetic field \(\mathbf{B}=\mathrm{B}_0 \hat{\mathbf{i}}\), each with an equal momentum of magnitude \(p=e B R\). Under what conditions on the direction of momentum will the orbits be nonintersecting circles?

Answer: The magnetic field \(B\) is along the \(x\)-axis, hence for a circular orbit the momenta of the two particles are in the \(y-z\) plane. Let \(p_1\) and \(p_2\) be the momentum of the electron ( \(e^{-}\)) and positron ( \(e^{+}\)), respectively. Both traverse a circle of radius \(R\) of opposite sense. Let \(p\), make an angle \(\theta\) with they-axis \(p_2\) must make the same angle withy axis.
The centres of the respective circles must be perpendicular to the momenta and at a distance R. Let the centre of the electron be at \(C_e\) and of the positron at \(C_p\).
The coordinates of \(C_e\) is \(C_e=(0,-R \sin \theta, R \cos \theta)\)
The coordinates of \(C_p\) is \(C_p=[0,-R \sin \theta,(1.5 R-R \cos \theta)]\)

The circular orbits of electron and positron shall not overlap if the distance between the two centers are greater than \(2 R\).
Let \(d\) be the distance between \(C_p\) and \(C_e\). Then,
\(
\begin{aligned}
d^2 & =[R \sin \theta-(-R \sin \theta)]^2+\left[R \cos \theta-\left(\frac{3}{2} R-R \cos \theta\right)\right]^2 \\
& =(2 R \sin \theta)^2+\left(2 R \cos \theta-\frac{3}{2} R\right)^2 \\
& =4 R^2 \sin ^2 \theta+4 R^2 \cos ^2 \theta-6 R^2 \cos \theta+\frac{9}{4} R^2 \\
& =4 R^2+\frac{9}{4} R^2-6 R^2 \cos \theta
\end{aligned}
\)
As, \(d\) has to be greater than \(2 R, d^2>4 R^2\)
\(
\Rightarrow \quad 4 R^2+\frac{9}{4} R^2-6 R^2 \cos \theta>4 R^2
\)
or, \(\frac{9}{4}>6 \cos \theta\) or, \(\cos \theta<\frac{3}{8}\)

Q4.26: A uniform conducting wire of length \(12 a\) and resistance \(R\) is wound up as a current carrying coil in the shape of (i) an equilateral triangle of side \(a\); (ii) a square of sides \(a\) and, (iii) a regular hexagon of sides \(a\). The coil is connected to a voltage source \(V_0\). Find the magnetic moment of the coils in each case.

Answer: Key concept: In this problem different shapes form figures of different area and the number of loops in each case is different and hence, there magnetic moments varies.
Magnetic moment is \(\mathrm{m}=\mathrm{nIA}\).
Since, the same wire is used in three cases with same potentials, therefore, same current flows in three cases.

(i) For an equilateral triangle of side \({a}\),
As the total wire of length \(=12 a\), so, the no. of loops \(n=\frac{12 a}{3 a}=4\)
Magnetic moment of the coils \(m=n I A\)
\(
\begin{aligned}
& \text { As area of triangle is } A=\frac{\sqrt{3}}{4} a^2 \\
& =4 I\left(\frac{\sqrt{3}}{4} a^2\right) \\
& \therefore \quad m=I a^2 \sqrt{3}
\end{aligned}
\)
(ii) For a square of sides \(a\),
\(
A=a^2
\)
No. of loops \(n=\frac{12 a}{4 a}=3\)
Magnetic moment of the coils \(m=n I A=3 I\left(a^2\right)=3 I a^2\)
(iii) For a regular hexagon of sides \(a\),
No. of loops \(n=\frac{12 a}{6 \dot{a}}=2\)
Area, \(\quad A=\frac{6 \sqrt{3}}{4} a^2\)
Magnetic moment of the coils \(m=n I A\)
\(
\Rightarrow \quad m=2 I\left(\frac{6 \sqrt{3}}{4} a^2\right)
\)
\(\Rightarrow \quad m=3 \sqrt{3} a^2 I, m\) is in a geometric series.

Q4.27: Consider a circular current-carrying loop of radius \(R\) in the \(x-y\) plane with centre at origin. Consider the line intergral
\(
\mathfrak{J}(L)=\left|\int_{-L}^L \mathbf{B} . \mathbf{d l}\right| \text { taken along z-axis. }
\)
(a) Show that \(\mathfrak{J}(L)\) monotonically increases with \(L\).
(b) Use an appropriate Amperian loop to show that \(\mathfrak{J}(\infty)=\mu_0 I\), where \(I\) is the current in the wire.
(c) Verify directly the above result.
(d) Suppose we replace the circular coil by a square coil of sides \(R\) carrying the same current \(I\). What can you say about \(\mathfrak{J}(L)\) and \(\mathfrak{J}(\infty)\) ?

Answer: 

(a) Magnetic field due to a circular current-carrying loop lying in the \(x y\) – plane acts along \(z\)-axis as shown in figure.
\(\mathrm{B}(\mathrm{z})\) point in the same direction of \(Z\)-axis and hence \(\mathfrak{J}(L)\) is monotonical function of \(L\) as
\(B\) and \(dl\) are along the same direction. So
\(
B \cdot d l=B d l \cos \theta=B d l \cos 0^{\circ}=B d l .
\)
\(\therefore \mathfrak{J}(L)\) is monotonically increasing function of \(L\).
(b) Now consider an Amperean loop around the circular coil of such a large radius that \(L \rightarrow \infty\). Since this loop encloses a current \(I\), Now using Ampere’s law
\(
\mathfrak{J}(\infty)=\oint_{-\infty}^{+\infty} \vec{B} \cdot \overrightarrow{d l}=\mu_0 I
\)
(c) The magnetic field at the axis (z-axis) of circular coil is given by \(B_z=\frac{\mu_0 I R^2}{2\left(z^2+R^2\right)^{3 / 2}}\)
Now integrating
\(
\int_{-\infty}^{+\infty} B_z d z=\int_{-\infty}^{+\infty} \frac{\mu_0 I {R}^2}{2\left(z^2+R^2\right)^{3 / 2}} d z
\)
Let \(z=R \tan \theta\) so that \(d z=R \sec ^2 \theta d \theta\)
and
\(
\begin{aligned}
\left(z^2+R^2\right)^{3 / 2} & =\left(R^2 \tan ^2 \theta+R^2\right)^{3 / 2} \\
& =R^3 \sec ^3 \theta \quad\left(\text { as } 1+\tan ^2 \theta=\sec ^2 \theta\right)
\end{aligned}
\)
Thus,
\(
\begin{aligned}
\int_{-\infty}^{+\infty} B_z d z & =\frac{\mu_0 I}{2} \int_{-\pi / 2}^{+\pi / 2} \frac{R^2\left(R \sec ^2 \theta d \theta\right)}{R^3 \sec ^3 \theta} \\
& =\frac{\mu_0 I}{2} \int_{-\pi / 2}^{+\pi / 2} \cos \theta d \theta=\mu_0 I
\end{aligned}
\)
(d) As we know \(\left(B_z\right)_{\text {square }}<\left(B_z\right)_{\text {circular coil }}\)
For the same current, and side of the square equal to radius of the coil
\(\mathfrak{J}(\infty)_{\text{square}}< \mathfrak{J}(\infty)_{\text{circular coil}}\)
By using the same argument as we done in case (b), it can be shown that
\(\mathfrak{J}(\infty)_{\text{square}} = \mathfrak{J}(\infty)_{\text{circular coil}}\)

Q4.28: A multirange current meter can be constructed by using a galvanometer circuit as shown in Fig. 4.5. We want a current meter that can measure \(10 \mathrm{~mA}, 100 \mathrm{~mA}\) and 1 A using a galvanometer of resistance \(10 \Omega\) and that prduces maximum deflection for current of 1 mA . Find \(S_1, S_2\) and \(S_3\) that have to be used.

Answer: Key concept: A galvanometer can be converted into ammeter by connecting a very low resistance wire (shunt \(S\) ) connected in parallel with galvanometer. The relationship is given by \(\mathrm{I}_g G=\left(I-I_g\right) \mathrm{S}\), where \(\mathrm{I}_g\) is the range of galvanometer, \(G\) is the resistance of galvanometer.

For measuring \(I_1=10 \mathrm{~mA}: I_G \cdot G=\left(I_1-I_G\right)\left(S_1+S_2+S_3\right)\)
For measuring \(I_2=100 \mathrm{~mA}: I_G\left(G+S_1\right)=\left(I_2-I_G\right)\left(S_2+S_3\right)\)
For measuring \(I_3=1 \mathrm{~A}: I_G\left(G+S_1+S_2\right)=\left(I_3-I_G\right)\left(S_3\right)\)
gives \(\quad S_1=1 \Omega, S_2=0.1 \Omega\)
and \(\quad S_3=0.01 \Omega\)

Q4.29: Five long wires A, B, C, D and E, each carrying current \(I\) are arranged to form edges of a pentagonal prism as shown in Fig. 4.6. Each carries current out of the plane of paper.
(a) What will be magnetic induction at a point on the axis O? Axis is at a distance \(R\) from each wire.
(b) What will be the field if current in one of the wires (say A) is switched off?
(c) What if current in one of the wire (say) \(A\) is reversed?

Answer: (a) Key concept: The wires shown in this problem carrying current outwards to the plane. And we know that direction of magnetic field is perpendicular to both current and position vector \(r\) . So, the vector sum of magnetic field produced by each wire at 0 is equal to 0 .
Suppose the five wires A, B, C, D and E be perpendicular to the plane of paper at locations as shown in figure.
Thus, magnetic field induction due to five wires will be represented by various sides of closed pentagon in one order, lying in the plane of paper. So, its value is zero.
(b) When current in \(\mathrm{AA}^{\prime}\) is switched off, then \(\mathrm{B}_1=0\) and resultant becomes
\(
\mathrm{R}=\mathrm{B}_2+\mathrm{B}_3+\mathrm{B}_4+\mathrm{B}_5
\)
But from (a) part \(\mathrm{B}_1+\mathrm{B}_2+\mathrm{B}_3+\mathrm{B}_4+\mathrm{B}_5=0\)
Or \(\vec{B}_2+\vec{B}_3+\vec{B}_4+\vec{B}_5=-\vec{B}_1\)
\(
\mathrm{R}=-\mathrm{B}_1
\)
\(
R=\frac{\mu_0 I}{2 \pi r}
\)
i.e. direction of resultant is opposite to \(\vec{B}_1\)
(c) If current in wire \(A\) is reversed, then
Total magnetic field induction at \(O\)
\(=\) Magnetic field induction due to wire \(A+\) Magnetic field induction due to wires \(B, C, D\) and \(E\)
\(
=\frac{\mu_0}{4 \pi R} \frac{2 I}{R} \text { (acting perpendicular to } A O \text { towards left) }+\frac{\mu_0}{\pi} \frac{2 I}{R}
\)
(acting perpendicular \(A O\) towards left) \(=\frac{\mu_0 I}{\pi R}\) acting perpendicular \(A O\) towards left.