Past NEET Entrance Papers

Hint

(b) 

\(
\begin{aligned}
&\text { The magnitude of magnetic field due to circular coil of } \mathrm{N} \text { turns is given by }\\
&\begin{aligned}
& B_C=\frac{\mu_0 i N}{2 R} \\
& =\frac{4 \pi \times 10^{-7} \times 7 \times 100}{2 \times 0.1} \\
& =4.4 \times 10^{-3} \mathrm{~T} \\
& =4.4 \mathrm{mT}
\end{aligned}
\end{aligned}
\)

Hint

(b) According to modified Ampere’s law
\(
\oint B \cdot d I=\mu_0\left(I_C+I_D\right)
\)
For Loop \(L_1 \quad I_C \neq 0\) and \(I_D=0\)
For Loop \(L_2 \quad I_C=0\) and \(I_D \neq 0\)
Due to KCL \(\quad I_C=I_D\)

Hint

(d)

\(
\begin{aligned}
&\mathrm{mg}=\mathrm{ilB}\\
&\begin{gathered}
250 \times 10^{-3} \times 9.8=i \times 2 \times 0.7 \\
\Rightarrow i=\frac{0.250 \times 9.8}{2 \times 0.7} \\
\Rightarrow i=1.75 \mathrm{~A}
\end{gathered}
\end{aligned}
\)

Hint

Step 1: Understand the scenario
We have a uniform electric field (E) and a uniform magnetic field
(B) acting in the same direction. An electron is projected along the direction of these fields with a certain velocity (V).
Hint: Visualize the setup by drawing the electric field, magnetic field, and the direction of the electron’s velocity.
Step 2: Analyze the forces acting on the electron
When an electron is placed in an electric field, it experiences an electric force given by:
\(
F_E=q E
\)
where \(q\) is the charge of the electron (which is negative).
In a magnetic field, the force on a charged particle is given by:
\(
F_B=q(\mathbf{V} \times \mathbf{B})
\)
Since the electron is moving in the same direction as the magnetic field, the cross product \(\mathbf{V} \times \mathbf{B}\) will be zero:
\(
F_B=0
\)
Hint: Remember that the magnetic force depends on the angle between the velocity and the magnetic field. If they are parallel, the magnetic force is zero.
Step 3: Determine the net force acting on the electron
Since the magnetic force is zero, the only force acting on the electron is the electric force. The direction of the electric force is opposite to the direction of the electric field (since the electron has a negative charge).
Hint: Identify the direction of the electric force relative to the velocity of the electron.
Step 4: Analyze the effect of the electric force
The electric force will cause the electron to decelerate because it acts in the opposite direction to the velocity of the electron. This means that the speed of the electron will decrease over time.
Hint: Consider how forces affect motion according to Newton’s second law.
Step 5: Conclusion
Since the electron is experiencing a force that opposes its motion, its speed will decrease. Therefore, the correct conclusion is that the speed of the electron reduces.
Final Answer: The speed of the electron decreases.

Hint

(d) The magnetic force \(\vec{F}\) on a current-carrying wire in a magnetic field is given by:
\(
\vec{F}=I \vec{L} \times \vec{B}
\)
where \(\vec{L}\) is the vector representing the length and direction of the wire. Since the wire is along the positive X -axis, we can represent it as:
\(
\vec{L}=L \hat{i}
\)
Now we need to calculate the cross product \(\vec{L} \times \vec{B}\) :
\(
\vec{F}=I(L \hat{i}) \times(2 \hat{i}+3 \hat{j}-4 \hat{k})
\)
Using the determinant form for the cross product:
\(
\vec{F}=I L\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 0 & 0 \\
2 & 3 & -4
\end{array}\right|
\)
Expanding the determinant:
\(
\begin{gathered}
\vec{F}=I L(\hat{i}(0 \cdot-4-0 \cdot 3)-\hat{j}(1 \cdot-4-0 \cdot 2) \\
+\hat{k}(1 \cdot 3-0 \cdot 2))
\end{gathered}
\)
Simplifying this:
\(
\vec{F}=I L(0 \hat{i}+4 \hat{j}+3 \hat{k})
\)
Therefore, we have:
\(
\vec{F}=I L(4 \hat{j}+3 \hat{k})
\)
\(
\begin{aligned}
&\text { The magnitude of the force }|\vec{F}| \text { is given by: }\\
&\begin{gathered}
|\vec{F}|=I L \sqrt{(4)^2+(3)^2}=I L \sqrt{16+9}=I L \sqrt{25} \\
=5 I L
\end{gathered}
\end{aligned}
\)

Hint

(d) 

\(
\begin{aligned}
& \mathrm{B}=\frac{\mu_0}{4 \pi} \frac{\mathrm{i}}{\mathrm{R}}(\pi)-\frac{\mu_0}{4 \pi} \frac{2 \mathrm{i}}{\mathrm{R}} \\
& =\frac{\mu_0 \mathrm{i}}{4 \mathrm{R}}\left[1-\frac{2}{\pi}\right] \text { outward i.e. away from page. }
\end{aligned}
\)

Hint

(c)

\(
B=\mu_0 n i=4 \pi \times 10^{-7} \times \frac{100}{10^{-3}} \times 1=12.56 \times 10^{-2} T
\)

Hint

(a) Magnetic field at the centre of coil
\(
B=\frac{\mu_0 N I}{2 R}
\)
On substituting the given values
\(
\begin{aligned}
& \Rightarrow B=\frac{4 \pi \times 10^{-7} \times 10^3}{2 \times 62.8 \times 10^{-2}} \\
& =10^{-3} \mathrm{~T}
\end{aligned}
\)

Hint

(c) From the right hand curl rule, the shape of magnetic field lines due to long current carrying wire is circular (concentric circles).

Hint

(d)

Force per unit length between two long current carrying wires \(=\frac{\mu_0 I_1 I_2}{2 \pi d}\)
\(
F=\frac{4 \pi \times 10^{-7} \times 5 \times 10}{2 \pi \times 10 \times 10^{-2}}=10^{-4} \mathrm{Nm}^{-1}
\)
\(\Rightarrow\) The force is attractive as direction of current in two conductors is same
Hence, required force is \(1 \times 10^{-4} \mathrm{Nm}^{-1}\) and it is attractive.

Hint

(b)

\(
\begin{aligned}
&\text { Magnetic field at the axis of a current carrying circular loop }\\
&|B|=\frac{\mu_0 i a^2}{2\left[a^2+d^2\right]^{3 / 2}} \Rightarrow|B|=\frac{4 \pi \times 10^{-7} \times \sqrt{2} \times 1}{2[1+1]^{3 / 2}} \\
=3.14 \times 10^{-7} \mathrm{~T}
\end{aligned}
\)

Hint

(d) According to Biot-Savart’s law \(d \vec{B}=\frac{\mu_0}{4 \pi} \frac{I \overrightarrow{d l} \times \vec{r}}{r^3}\) which is applicable for infinitesimal element. It is analogous to Coulomb’s law, where \(I \overrightarrow{d l}\) is vector source and electric field is produced by scalar source q. Here statement I is correct and statement II is incorrect.

Hint

(c) For solid wire

Inside point:
\(
\begin{aligned}
& B=\frac{\mu_0 I r^2}{R^2 \times 2 \pi r} \\
& =\frac{\mu_0 I r}{R^2 \times 2 \pi} \\
& B \propto r
\end{aligned}
\)

Outside point:
\(
\begin{aligned}
& B=\frac{\mu_0 I}{2 \pi r} \\
& B \propto \frac{1}{r}
\end{aligned}
\)

Hint

(a)

The ratio of currents in the two coils is \(4: 1\). This means that the current in the first coil should be four times the current in the second coil to produce the same magnetic moment.

Explanation
(1) Define the variables. Let the radii of the two coils be \(r\) and \(2 r\), respectively. Let \(I_1\) and \(I_2\) be the currents flowing through the coils with radii \(r\) and \(2 r\), respectively.
(2) Express the magnetic moment for each coil. The magnetic moment of a circular coil is given by \(M=I A\), where \(I\) is the current and \(A\) is the area of the coil. For the first coil, the magnetic moment is \(M_1=I_1 \pi r^2\). For the second coil, the magnetic moment is \(M_2=I_2 \pi(2 r)^2=4 I_2 \pi r^2\).
(3) Equate the magnetic moments. Since the magnetic moment is the same for both coils, we have \(M_1=M_2\). Therefore, \(I_1 \pi r^2=4 I_2 \pi r^2\).
(4) Solve for the ratio of currents. Canceling out the common terms, we get \(I_1=4 I_2\). Therefore, the ratio of currents is \(\frac{I_1}{I_2}=\frac{4 I_2}{I_2}=4: 1\).

Hint

(b) If a strong magnetic field is applied along the direction of an electron’s velocity, the electron would continue to move along the same path, experiencing no deflection, as the magnetic force acting on it would be zero due to the parallel alignment of the velocity and magnetic field.

Hint

(a) \(
\begin{aligned}
&\text { From Ampere’s circuital law }\\
&\begin{aligned}
& B_{i n}=\frac{\mu_0 I r}{2 \pi R^2} \\
& B_0=\frac{\mu_0 I}{2 \pi r}
\end{aligned}
\end{aligned}
\)

Hint

(d)

\(
\begin{aligned}
& B=\frac{\mu_0 I}{2 \pi R} \\
& \mathrm{~F}=\mathrm{BVq} \sin \theta \\
& \theta=90^{\circ} \\
& \mathrm{F}=\mathrm{BVq} \\
& F=\frac{\mu_0 I}{2 \pi R} \times V \times e=\frac{2 \times 10^{-7} \times 5}{20 \times 10^{-2}} \times 10^5 \times 1.6 \times 10^{-19} \\
& \mathrm{~F}=8 \times 10^{-20} \mathrm{~N}
\end{aligned}
\)

Hint

(c)
(i) If it is an equilateral triangle of side ‘ \(a\) ‘ no. of triangular loops formed \(\frac{12 a}{3 a}=4\) and magnetic moment \(4 \cdot \frac{\sqrt{3} a^2}{4} l=\sqrt{3} l a^2\)
(ii) Side of the square if it is a.
No. of square formed \(\frac{12 a}{4 a}=3\)
Magnetic moment of the square loop is \(3 a^2 I\).

Hint

(d) The magnetic force is given by:
\(
\vec{F}=q(\vec{v} \times \vec{B})
\)
Since \(q=1\), we have:
\(
\vec{F}=\vec{v} \times \vec{B}
\)
\(
\vec{v} \times \vec{B}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 4 & 6 \\
B & B & B_0
\end{array}\right|
\)
\(
\vec{v} \times \vec{B}=\left(4 B_0-6 B\right) \hat{i}-\left(2 B_0-6 B\right) \hat{j}
\)
\(
\begin{aligned}
&\text { Now, we set this equal to the force vector: }\\
&\left(4 B_0-6 B\right) \hat{i}-\left(2 B_0-6 B\right) \hat{j}=4 \hat{i}-20 \hat{j}+12 \hat{k}
\end{aligned}
\)
From the \(\hat{\boldsymbol{i}}\) component:
\(
4 B_0-6 B=4 \dots(1)
\)
From the \(\hat{\boldsymbol{j}}\) component:
\(
-\left(2 B_0-6 B\right)=-20 \Longrightarrow 2 B_0-6 B=20 \dots(2)
\)
The \(\hat{k}\) component gives no information since it equals zero.
From equation (1):
\(
\begin{gathered}
4 B_0-6 B=4 \Longrightarrow 4 B_0=6 B+4 \Longrightarrow B_0 \\
=\frac{6 B+4}{4}=\frac{3 B}{2}+1 \dots(3)
\end{gathered}
\)
From equation (2):
\(
\begin{gathered}
2 B_0-6 B=20 \Longrightarrow 2 B_0=6 B+20 \Longrightarrow B_0 \\
=3 B+10 \dots(4)
\end{gathered}
\)
Set equations (3) and (4) equal
\(
\frac{3 B}{2}+1=3 B+10
\)
Multiplying through by 2 to eliminate the fraction:
\(
3 B+2=6 B+20
\)
Rearranging gives:
\(
3 B=-18 \Longrightarrow B=-6
\)
Substitute \(B\) back to find \(B_0\)
Substituting \(B=-6\) into equation(3):
\(
B_0=\frac{3(-6)}{2}+1=-9+1=-8
\)
Final Result
Thus, the complete expression for \(\vec{B}\) is:
\(
\vec{B}=-6 \hat{i}-6 \hat{j}-8 \hat{k}
\)

Hint

(d) Magnetic field at the centre of solenoid,
\(
B_{\text {solenoid }}=\mu_0 n l
\)
Given : No. of turns / length,
\(
n=\frac{N}{L}=\frac{100}{50 \times 10^{-2}}=200 \text { turns } / \mathrm{m}
\)
Current, \(I=2.5 \mathrm{~A}\)
\(
\begin{aligned}
\therefore B_{\text {solenoid }} & =\mu_0 n I=4 \pi \times 10^{-7} \times 200 \times 2.5 \\
& =6.28 \times 10^{-4} \mathrm{~T}
\end{aligned}
\)

Hint

(d) Step 1: Understand the Formula for Magnetic Moment
The magnetic moment \(M\) of a current-carrying loop is given by the formula:
\(
M=N \cdot I \cdot A
\)
where:
\(N\) is the number of turns (which is 1 for a single loop),
\(I\) is the current in amperes,
\(A\) is the area enclosed by the loop.
Step 2: Determine the Area of the Circle
Since the wire is bent into a circle, the area \(A\) of the circle can be calculated using the formula:
\(
A=\pi r^2
\)
where \(r\) is the radius of the circle.
Step 3: Relate the Length of the Wire to the Radius
The length of the wire \(l\) is equal to the circumference of the circle formed:
\(
l=2 \pi r
\)
From this, we can solve for the radius \(r\) :
\(
r=\frac{l}{2 \pi}
\)
Step 4: Substitute the Radius into the Area Formula
Now, substitute \(r\) back into the area formula:
\(
A=\pi\left(\frac{l}{2 \pi}\right)^2
\)
Calculating this gives:
\(
A=\pi \cdot \frac{l^2}{4 \pi^2}=\frac{l^2}{4 \pi}
\)
Step 5: Substitute the Area into the Magnetic Moment Formula
Now we can substitute the area \(A\) back into the magnetic moment formula:
\(
M=N \cdot I \cdot A=1 \cdot I \cdot \frac{l^2}{4 \pi}
\)
Thus, the magnetic moment is:
\(
M=\frac{I l^2}{4 \pi}
\)
Step 6: Final Result
The magnetic moment of the wire bent in the form of a circle is:
\(
M=\frac{I l^2}{4 \pi}
\)

Hint

(d) Inside \((\mathrm{d}<\mathrm{R})\)
Magnetic field inside conductor
\(
B=\frac{\mu_0}{2 \pi} \frac{i}{R^2} d
\)
\(\text { or } B=K d \ldots \text { (i) }\)
Straight line passing through origin
At surface ( \(\mathrm{d}=\mathrm{R}\) )
\(B=\frac{\mu_0}{2 \pi} \frac{i}{R} \dots(ii)\) Maximum at surface
\(
\text { Outside ( } d>R \text { ) }
\)
\(
\begin{aligned}
&B=\frac{\mu_0}{2 \pi} \frac{i}{d}\\
&\text { or } B \propto \frac{1}{d} \text { (Hyperbolic) }
\end{aligned}
\)

Hint

(b) Radius of the path \(=r=\frac{m v}{q B}=\frac{P}{q B}\)
For \(\mathrm{H}^{+}\)ion, \(r_H=\frac{p_H}{e B}\)
For \(\alpha\) particle
\(
\begin{aligned}
& r_\alpha=\frac{p_\alpha}{2 e B} \\
& \frac{r_H}{r_\alpha}=\frac{\frac{p}{e B}}{\frac{p}{2 e B}} \\
& \Rightarrow \frac{r_H}{r_\alpha}=\frac{2}{1}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { (a) } B=\frac{\mu_0 \mathrm{~N} \cdot I}{2 \pi R} \\
& \therefore \frac{B_1}{B_2}=\frac{N_1 R_2}{N_2 R_1}=\frac{200}{100} \frac{20}{40}=1 \\
& \text { So, } \frac{B_1}{B_2}=1
\end{aligned}
\)

Hint

(a) Magnetic field due to \(i_1=\frac{\mu_0 i_1}{2 R} \frac{\theta_1}{2 \pi}\) (Into the plane)
Magnetic field due to \(i_2=\frac{\mu_1 i_2}{2 R} \frac{\theta_2}{2 \pi}\) (out of the plane )
For parallel combination \(\frac{i_1}{i_2}=\frac{\rho l_2}{A} \times \frac{A}{\rho l_1}=\frac{l_1}{l_2}\)
\(
\begin{aligned}
& \Rightarrow \frac{i_1}{i_2}=\frac{\frac{1}{1}(2 \pi R)}{\frac{3}{4}(2 \pi R)} \\
& =\frac{1}{3} \\
& \Rightarrow i_1=\frac{i_2}{3} \\
& \Rightarrow i_2=3 i_1
\end{aligned}
\)
\(\therefore\) Net magnetic field
\(
\begin{aligned}
& =\frac{\mu_0 i_1}{2 R}\left(\frac{\theta_1}{2 \pi}\right)-\frac{\mu_0 i_2}{2 R}\left(\frac{\theta_2}{2 \pi}\right) \\
& =\frac{\mu_0}{2 R}\left(\frac{3 \pi}{2 \times 2 \pi}\right)-\frac{\mu_0 i_2}{2 R}\left(\frac{\pi}{2 \times 2 \pi}\right) \\
& =\frac{\mu_0}{2 R}\left[\frac{3 i_1}{4}-\frac{i_2}{4}\right] \\
& =\frac{\mu_0}{2 R}\left[\frac{3 i_1}{4}-\frac{3 i_1}{4}\right]=0
\end{aligned}
\)

Hint

(c)

From figure, for equilibrium, \(\mathrm{mg} \sin 30^{\circ}=\mathrm{I} \ell \mathrm{B} \cos 30^{\circ}\)
\(
\begin{aligned}
\Rightarrow \quad \mathrm{I} & =\frac{m g}{\ell B} \tan 30^{\circ} \\
& =\frac{0.5 \times 9.8}{0.25 \times \sqrt{3}}=11.32 \mathrm{~A}
\end{aligned}
\)

Hint

(c) Current sensitivity of moving coil galvanometer
\(
\mathrm{I}_{\mathrm{s}}=\frac{N B A}{C} \dots(i)
\)
Voltage sensitivity of moving coil galvanometer,
\(
\mathrm{V}_{\mathrm{s}}=\frac{N B A}{C R_G} \dots(ii)
\)
Dividing eqn. (i) by (ii)
Resistance of galvanometer
\(
\mathrm{R}_{\mathrm{G}}=\frac{I_s}{V_s}=\frac{5 \times 1}{20 \times 10^{-3}}=\frac{5000}{20}=250 \Omega
\)

Hint

(d) Work done, \(\mathrm{W}=\mathrm{MB}\left(\cos \theta_1-\cos \theta_2\right)\)
When it is rotated by angle \(180^{\circ}\) then
\(
\begin{aligned}
& \mathrm{W}=\mathrm{MB}\left(\cos 0^{\circ}-\cos 180^{\circ}\right)=\mathrm{MB}(1+1) \\
& \mathrm{W}=2 \mathrm{MB} \\
& \mathrm{~W}=2(\mathrm{NIA}) \mathrm{B} \\
& =2 \times 250 \times 85 \times 10^{-6}\left[1.25 \times 2.1 \times 10^{-4}\right] \times 85 \times 10^{-2} \\
& =9.1 \mu \mathrm{~J}
\end{aligned}
\)

Hint

(d)

Force per unit length between two parallel current carrying conductors,
\(
\mathrm{F}=\frac{\mu_0 \mathrm{i}_1 \mathrm{i}_2}{2 \pi \mathrm{~d}}
\)
Since same current flowing through both the wires
\(
\begin{aligned}
& \mathrm{i}_{\mathrm{i}}=\mathrm{i}_2=\mathrm{i} \\
& \text { so } \mathrm{F}_1=\frac{\mu_0 \mathrm{i}^2}{2 \pi \mathrm{~d}}=\mathrm{F}_2
\end{aligned}
\)
\(\therefore\) Magnitude of force per unit length on the middle wire ‘ B ‘
\(
\mathrm{F}_{\mathrm{net}}=\sqrt{\mathrm{F}_1^2+\mathrm{F}_2^2}=\frac{\mu_0 \mathrm{i}^2}{\sqrt{2} \pi \mathrm{~d}}
\)

Hint

(a) Here. \(B=3.57 \times 10^{-2} \mathrm{~T}\),
\(
\frac{e}{m}=1.76 \times 10^{11} \mathrm{C} \mathrm{~kg}^{-1}
\)
Frequency of revolution of the electron,
\(
\nu=\frac{1}{T}=\frac{v}{2 \pi r} \dots(i)
\)
Also, \(\frac{m v^2}{r}=e v B \Rightarrow \frac{v}{r}=\frac{e B}{m} \dots(ii)\)
From eqns. (i) and (ii),
\(
\begin{aligned}
\nu =\frac{1}{2 \pi} \times \frac{e B}{m}=\frac{1}{2 \times 3.14} \times 1.76 \times 10^{11} \times 3.57 \times 10^{-2} \\
& =10^9 \mathrm{~Hz}=1 \mathrm{GHz}
\end{aligned}
\)

Hint

(a)

The direction of current in conductor
\(
\begin{aligned}
& \mathrm{XY} \text { and } \mathrm{AB} \text { is same } \\
& \therefore \quad \mathrm{F}_{\mathrm{AB}}=\mathrm{i} \ell \mathrm{~B} \quad \text { (attractive) } \\
& \mathrm{F}_{\mathrm{AB}}=\mathrm{i}(\mathrm{~L}) \cdot \frac{\mu_0 \mathrm{I}}{2 \pi\left(\frac{\mathrm{~L}}{2}\right)}(\leftarrow)=\frac{\mu_0 \mathrm{iI}}{\pi}(\leftarrow)
\end{aligned}
\)
\(
\begin{aligned}
& \mathrm{F}_{\mathrm{BC}} \text { opposite to } \mathrm{F}_{\mathrm{AD}} \\
& \mathrm{~F}_{\mathrm{BC}}(\uparrow) \text { and } \mathrm{F}_{\mathrm{AD}}(\downarrow) \\
& \Rightarrow \text { cancels each other } \\
& \mathrm{F}_{\mathrm{CD}}=\mathrm{i} \ell \mathrm{B} \text { (repulsive) }
\end{aligned}
\)
\(
\mathrm{F}_{\mathrm{CD}}=\mathrm{i}(\mathrm{~L}) \frac{\mu_0 \mathrm{I}}{2 \pi\left(\frac{3 \mathrm{~L}}{2}\right)}(\rightarrow)=\frac{\mu_0 \mathrm{iI}}{3 \pi}(\rightarrow)
\)
Therefore the net force on the loop
\(
\mathrm{F}_{\mathrm{net}}=\mathrm{F}_{\mathrm{AB}}+\mathrm{F}_{\mathrm{BC}}+\mathrm{F}_{\mathrm{CD}}+\mathrm{F}_{\mathrm{AD}}
\)
\(
\Rightarrow \quad \mathrm{F}_{\mathrm{net}}=\frac{\mu_{\mathrm{o}} \mathrm{iI}}{\pi}-\frac{\mu_{\mathrm{o}} \mathrm{iI}}{3 \pi}=\frac{2 \mu_{\mathrm{o}} \mathrm{iI}}{3 \pi}
\)

Hint

(a) For One turn loop :
\(
\mathrm{B}=\frac{\mu_0 i}{2 r}
\)
For \(n\) turn loop :
\(
\begin{aligned}
& \mathrm{r}=\frac{R}{n} \\
& \mathrm{~B}^{\prime}=\frac{\mu_0 n i}{2 \frac{R}{n}} \\
& =\left(\frac{\mu_0 i}{2 R}\right) n^2 \\
& ={n}^2 \mathrm{~B}
\end{aligned}
\)

Alternate:

(a) Let \(l\) be the length of the wire. Magnetic field at the centre of the loop is
\(
\begin{aligned}
& B=\frac{\mu_0 I}{2 R} \\
\therefore & B=\frac{\mu_0 \pi I}{l} (\because l=2 \pi R) \dots(i)\\
B^{\prime}= & \frac{\mu_0 n I}{2 r}=\frac{\mu_0 n I}{2\left(\frac{l}{2 n \pi}\right)} \\
B^{\prime}= & \frac{\mu_0 n^2 \pi I}{l} \dots(ii)
\end{aligned}
\)
From eqns. (i) and (ii), we get
\(
B^{\prime}=n^2 B
\)

Hint

(b) Magnetic field at a point inside the wire at distance \(r\left(=\frac{a}{2}\right)\) from the axis of wire is
\(
B=\frac{\mu_0 I}{2 \pi a^2} r=\frac{\mu_0 I}{2 \pi a^2} \times \frac{a}{2}=\frac{\mu_0 I}{4 \pi a}
\)
Magnetic field at a point outside the wire at distance \(r(=2 a)\) from the axis of wire is
\(
B^{\prime}=\frac{\mu_0 I}{2 \pi r}=\frac{\mu_0 I}{2 \pi} \times \frac{1}{2 a}=\frac{\mu_0 I}{4 \pi a} \quad \therefore \frac{B}{B^{\prime}}=1
\)

Hint

(d) As we know, \(F=q v B=\frac{m v^2}{R}\)
\(
\therefore \mathrm{R}=\frac{m v}{q B}=\frac{\sqrt{2 m(k E)}}{q B}
\)
Since R is same so, \(\mathrm{KE} \propto \frac{q^2}{m}\) Therefore KE of \(\alpha\) particle
\(
=\frac{q^2}{m}=\frac{(2)^2}{4}=1 \mathrm{MeV}
\)

Hint

(a) The ammeter has a coil resistance of 480 ohms and a shunt of 20 ohms. Since the shunt is connected in parallel with the coil, we need to find the equivalent resistance of the ammeter.
The formula for the equivalent resistance \(R_{\text{s h u n t}}\) of two resistors in parallel \(R_1\) and \(R_2\) is given by:
\(
\frac{1}{R_{e q}}=\frac{1}{R_1}+\frac{1}{R_2}
\)
\(
\frac{1}{R_{e q}}=\frac{1}{480}+\frac{24}{480}=\frac{25}{480}
\)
\(
R_{e q}=\frac{480}{25}=19.2 \mathrm{ohm}
\)
Now, we need to find the total resistance in the circuit, which is the sum of the external resistor and the equivalent resistance of the ammeter:
\(
\begin{aligned}
& R_{\text {total }}=R_{\text {external }}+R_{e q}=40.8 \mathrm{ohm}+19.2 \mathrm{ohm} \\
&=60 \mathrm{ohm}
\end{aligned}
\)
\(
I=\frac{30 \mathrm{~V}}{60 \mathrm{ohm}}=0.5 \mathrm{~A}
\)

Hint

(b) To find the torque required to keep the rectangular coil in stable equilibrium, we can use the formula for torque \((\tau)\) on a current-carrying coil in a magnetic field:
\(
\tau=n \cdot I \cdot A \cdot B \cdot \sin (\theta)
\)
Where:
\(n=\) number of turns of the coil
\(I=\) current flowing through the coil
\(A=\) area of the coil
\(B=\) magnetic field strength
\(\theta=\) angle between the normal to the coil and the magnetic field
The area \(A\) of the rectangular coil can be calculated using the formula:
\(A=\) length \(\times\) width
Given:
Length \(=0.12 \mathrm{~m}\)
Width \(=0.1 \mathrm{~m}\)
Calculating the area:
\(
A=0.12 m \times 0.1 m=0.012 m^2
\)
We have:
Number of turns \(n=50\)
Current \(I=2 \mathrm{~A}\)
Magnetic field strength \(B=0.2\) Weber \(/ \mathrm{m}^2\)
Angle \(\theta=30^{\circ}\)
The angle \(\theta\) used in the torque formula is the angle between the magnetic field and the normal to the coil. Since the coil is inclined at \(30^{\circ}\) to the magnetic field, the angle between the normal to the coil and the magnetic field is:
\(
\theta=90^{\circ}-30^{\circ}=60^{\circ}
\)
\(
\begin{aligned}
&\text { Now we can substitute the values into the torque formula: }\\
&\tau=n \cdot I \cdot A \cdot B \cdot \sin \left(60^{\circ}\right)
\end{aligned}
\)
\(
\tau=50 \cdot 2 \cdot 0.012 \cdot 0.2 \cdot \frac{\sqrt{3}}{2}
\)
\(
\tau \approx 0.12 \cdot 1.732 \approx 0.2 \mathrm{~N} \mathrm{~m}
\)

Hint

(c)

Magnetic field due to segment ‘ 1 ‘
\(
\begin{aligned}
& \overrightarrow{B_1}=\frac{\mu_0 I}{4 \pi R}\left[\sin 90^{\circ}+\sin 0^{\circ}\right](-\widehat{k}) \\
& =\frac{-\mu_0 I}{4 \pi R}(\widehat{k})=\overrightarrow{B_3}
\end{aligned}
\)
Magnetic field due to segment 2
\(
\overrightarrow{B_2}=\frac{\mu_0 I}{4 R}(-\hat{i})=\frac{-\mu_0 I}{4 \pi R}(-\pi \hat{i})
\)
\(
\begin{aligned}
&\therefore \vec{B} \text { at centre }\\
&\begin{aligned}
& \overrightarrow{B_c}=\overrightarrow{B_1}+\overrightarrow{B_2}+\overrightarrow{B_3} \\
& =\frac{-\mu_0 I}{4 \pi R}(\pi \hat{i}+2 \widehat{k})
\end{aligned}
\end{aligned}
\)

Hint

(d) Current in the orbit, \(I=\frac{e}{T}\)
\(
I=\frac{e}{(2 \pi / \omega)}=\frac{\omega e}{2 \pi}=\frac{(2 \pi n) e}{2 \pi}=n e
\)
Magnetic field at centre of current carrying circular coil is given by
\(
B=\frac{\mu_0 I}{2 r}=\frac{\mu_0 n e}{2 r}
\)

Hint

(b)

The emf in AD:
\(
e_1=\left(a \times \mu_0 i v\right) / 2 \pi(x-a / 2)
\)
The emf in EF :
\(
e_2=\left(a \times \mu_0 i \times v\right) / 2 \pi(x+a / 2)
\)
Net emf \(=e_1-e_2\)
\(
\begin{aligned}
& \Rightarrow e=\frac{\mu_0}{2 \pi} a \times i \times v\left[\frac{1}{x-\frac{a}{2}}-\frac{1}{x+\frac{a}{2}}\right] \\
& e \propto \frac{1}{(2 x-a)(2 x+a)}
\end{aligned}
\)

Hint

(c) As \(0.2 \%\) of main current passes through the galvanometer hence \(\frac{998}{1000}\) I current through the shunt.

\(
\begin{aligned}
&I_G=0.2 \% \text { of } I=\frac{0.2}{100} I=\frac{1}{500} I\\
&\therefore \text { Current through the shunt, }\\
&I_S=I-I_G=I-\frac{1}{500} I=\frac{499}{500} I
\end{aligned}
\)
As shrunt and galvanometer are in parallel
\(
\begin{aligned}
& \therefore \quad I_G G=I_S S \\
& \quad\left(\frac{1}{500} I\right) G=\left(\frac{499}{500}\right) S \text { or } S=\frac{G}{499}
\end{aligned}
\)
Resistance of the ammeter \(R_A\) is
\(
\begin{aligned}
& \frac{1}{R_A}=\frac{1}{G}+\frac{1}{S}=\frac{1}{G}+\frac{1}{\frac{G}{499}}=\frac{500}{G} \\
& R_A=\frac{1}{500} G
\end{aligned}
\)

Hint

(d)

The magnetic field at the point \(P\), at a perpendicular distance \(d\) from \(O\) in a direction perpendicular to the plane \(A B C D\) due to currents through \(A O B\) and \(C O D\) are perpendicular to each other. Hence,
\(
\text { Net magnetic field, } B=\sqrt{B_1^2+B_2^2}
\)
\(
=\sqrt{\left(\frac{\mu_0 I_1}{2 \pi \mathrm{~d}}\right)^2+\left(\frac{\mu_0 \mathrm{I}_2}{2 \pi \mathrm{~d}}\right)^2}
\)
\(
\left(\because B_1=\frac{\mu_0 I_1}{2 \pi d} \text { and } B_2=\frac{\mu_0 I_2}{2 \pi d}\right)
\)
\(
=\frac{\mu_0}{2 \pi \mathrm{~d}} \sqrt{\mathrm{I}_1^2+\mathrm{I}_2^2}
\)

Hint

(c) A current loop in a magnetic field is in equilibrium in two orientations one is stable and another unstable.

\(
\begin{aligned}
& \because \quad \vec{\tau}=\vec{M} \times \vec{B}=\mathrm{M} \mathrm{~B} \sin \theta \\
& \text { If } \theta=0^{\circ} \Rightarrow \tau=0 \text { (stable) } \\
& \text { If } \theta=\pi \Rightarrow \tau=0 \text { (unstable) }
\end{aligned}
\)

Hint

(b)

When moves with an acceleration \(\mathrm{a}_0\) towards west, electric field
\(
\mathrm{E}=\frac{F}{q}=\frac{m a_0}{e} \text { (West) }
\)
When moves with an acceleration \(3 \mathrm{a}_0\) towards east, magnetic field
\(
\mathrm{B}=\frac{2 m a_0}{e v_0}(\text { downward })
\)

Hint

(b) Given:
Magnetic field \(B=2 \times 10^{-4}\) weber \(/ \mathrm{m}^2\)
Velocity of electron, \(v=10^7 \mathrm{~m} / \mathrm{s}\)
Lorentz force \(F=q \nu B \sin \theta \text { ( } \therefore \vec{v} \text { and } \vec{B} \text { are perpendicular to each other) }\)
\(
\begin{aligned}
& =1.6 \times 10^{-19} \times 10^7 \times 2 \times 10^{-4}\left(\because \theta=90^{\circ}\right) \\
& =3.2 \times 10^{-16} \mathrm{~N}
\end{aligned}
\)

Hint

(a) The net magnetic force on a current loop in a uniform magnetic field is always zero.
\(
\text { Here, } \vec{F}_{A B}+\vec{F}_{B C D A}=\overrightarrow{0}
\)
\(
\begin{aligned}
\Rightarrow & \vec{F}_{B C D A}=-\vec{F}_{A B}=-\vec{F} \\
& \left(\because F_{A B}=\vec{F}\right)
\end{aligned}
\)

Hint

(a)

Magnetic field induction due to verfical loop carrying current \(I\) ampere at the centre \(O\) is
\(
B_1=\frac{\mu_0 I}{2 R}
\)
It acts in horizontal direction.
The magnetic field due to the coil, carrying current \(2 I\) Ampere
\(
B_2=\frac{\mu_0(2 I)}{2 R}
\)
It acts in vertically upward direction.
As \(B_1\) and \(B_2\) are perpendicular to each other, therefore the resultant magnetic field induction at the centre \(O\) is
\(
\begin{aligned}
& B_{\text {net }}=\sqrt{B_1^2+B_2^2}=\sqrt{\left(\frac{\mu_0 I}{2 R}\right)^2+\left(\frac{\mu_0 2 I}{2 R}\right)^2} \\
& B_{\text {net }}=\frac{\mu_0 I}{2 R} \sqrt{(1)^2+(2)^2}=\frac{\sqrt{5} \mu_0 I}{2 R}
\end{aligned}
\)

Hint

(a) \(S=\frac{V_g}{\left(I-I_g\right)}\)
Neglecting \(I_g\)
\(
\therefore \quad S=\frac{V_g}{l}=\frac{25 \times 10^{-3} \mathrm{~V}}{25 \mathrm{~A}}=0.001 \Omega
\)

Hint

(a) According to the principal of circular motion in a magnetic field,
\(
\begin{aligned}
& F_c=F_m \Rightarrow \frac{m v^2}{R}=q V B \\
& \Rightarrow R=\frac{m v}{q B}=\frac{P}{q B}=\frac{\sqrt{2 m \cdot K}}{q B} \\
& R_\alpha=\frac{\sqrt{2(4 m) K^{\prime}}}{2 q B} \\
& \frac{R}{R_\alpha}=\sqrt{\frac{K}{K^{\prime}}} \\
& \text { but } R=R_\alpha \text { (given) } \\
& \text { Thus } K=K^{\prime}=1 \mathrm{MeV}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (c) Time period of cyclotron is }\\
&\begin{aligned}
& T=\frac{1}{\nu}=\frac{2 \pi m}{e B} ; B=\frac{2 \pi m}{e} \nu ; R=\frac{m \nu}{e B}=\frac{p}{e B} \\
& \Rightarrow \quad p=e B R=e \times \frac{2 \pi m \nu}{e} R=2 \pi m \nu R \\
& \text { K.E. }=\frac{p^2}{2 m}=\frac{(2 \pi m \nu R)^2}{2 m}=2 \pi^2 m \nu^2 R^2
\end{aligned}
\end{aligned}
\)

Hint

(b) Here, \(\vec{F}_{B C}=\vec{F}\)
\(
\because \quad \vec{F}_{A B}=0
\)
The net magnetic force on a current carrying closed loop in a uniform magnetic field is zero.
\(
\begin{aligned}
& \therefore \vec{F}_{A B}+\vec{F}_{B C}+\vec{F}_{A C}=0 \\
& \Rightarrow \vec{F}_{A C}=-\vec{F}_{B C}=-\vec{F} \quad\left(\because \vec{F}_{A B}=0\right)
\end{aligned}
\)

Hint

(a) Force on electron due to electric field,
\(
\vec{F}_E=-e \vec{E}
\)
Force on electron due to magnetic field,
\(
\vec{F}_B=-e(\vec{v} \times \vec{B})=0
\)
Since \(\vec{v}\) and \(\vec{B}\) are in the same direction.
Total force on the electron,
\(
\vec{F}=\vec{F}_E+\vec{F}_B=-e \vec{E}
\)
Electric field opposes the motion of the electron, hence speed of the electron will decrease.

Hint

(d)

To keep the main current in the circuit unchanged, the resistance of the galvanometer should be equal to the net resistance.
\(
\begin{aligned}
& \therefore G=\left(\frac{G S}{G+S}\right)+S^{\prime} \\
& \Rightarrow G-\frac{G S}{G+S}=S^{\prime} \\
& \therefore S^{\prime}=\frac{G^2}{G+S}
\end{aligned}
\)

Hint

(b) When the ring rotates about its axis with a uniform frequency \(f\) Hz , the current flowing in the ring is
\(
{I}=\frac{q}{T}={qf}
\)
Magnetic field at the centre of the ring is
\(
\mathrm{B}=\frac{\mu_0 I}{2 R}=\frac{\mu_0 q f}{2 R}
\)

Hint

(a)

\(\mathrm{F}_1>\mathrm{F}_2\) as \(F \propto \frac{1}{d}\), and \(\mathrm{F}_3\) and \(\mathrm{F}_4\) are equal and opposite. Hence, the net attraction force will be towards the conductor.

Hint

(a)

\(
\begin{aligned}
&\text { Magnetic fields due to the two parts at their common centre are respectively, }\\
&B_y=\frac{\mu_0 i}{4 R} \text { and } B_z=\frac{\mu_0 i}{4 R}
\end{aligned}
\)
\(
\begin{aligned}
& \text { Resultant field }=\sqrt{B_y^2+B_z^2} \\
& =\sqrt{\left(\frac{\mu_0 i}{4 R}\right)^2+\left(\frac{\mu_0 i}{4 R}\right)^2} \\
& =\sqrt{2} \frac{\mu_0 i}{4 R}=\frac{\mu_0 i}{2 \sqrt{2} R}
\end{aligned}
\)

Hint

(d)

\(
\begin{aligned}
&\text { Its the electron beam is not deflected then, }\\
&\begin{aligned}
& F_m=E e \\
& B e v=E e \\
& v=\frac{E}{B}
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
&\text { According to the law of conservation of energy }\\
&\begin{aligned}
& \frac{1}{2} m v^2=e V \\
& v=\sqrt{\frac{2 e V}{m}} \\
& \sqrt{\frac{2 e V}{m}}=\frac{E}{B} \\
& \frac{e}{m}=\frac{E^2}{2 V B^2}
\end{aligned}
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
&\text { Let the resistance to be added be } \mathrm{R} \text {, then }\\
&\begin{aligned}
& 30=\mathrm{I}_{\mathrm{g}}(\mathrm{r}+\mathrm{R}) \\
& \therefore \mathrm{R}=\frac{30}{I_g}-r \\
& =\frac{30}{30 \times 10^{-3}}-100 \\
& =1000-100=900 \Omega
\end{aligned}
\end{aligned}
\)

Hint

(b) When a current carrying loop is placed in a magnetic field, the coil experiences a torque given by \(\tau=N B i A \sin \theta\).
Torque is maximum when \(\theta=90^{\circ}\),i.e., the plane of the coil is parallel to the field \(\tau_{\max }=N B i A\)

Forces \(\vec{F}_1\) and \(\vec{F}_2\) acting on the coil are equal in magnitude and opposite in direction. As the forces \(\vec{F}_2\) and \(\vec{F}_2\) have the same line of action their resultant effect on the coil is zero. The two forces \(\vec{F}_3\) and \(\vec{F}_4\) are equal in magnitude and opposite in direction. As the two forces have different lines of action, they constitute a torque. Thus, if the force on one arc of the loop is \(\vec{F}\), the net force on the remaining three arms of the loop is \(-\vec{F}\).

Hint

(b) Magnetic moment of the loop.
\(
M=N I A=2000 \times 2 \times 1.5 \times 10^{-4}=0.6 \mathrm{~J} / \mathrm{T}
\)
Torque \(\tau=A B \sin 30^{\circ}\)
\(
=0.6 \times 5 \times 10^{-2} \times \frac{1}{2}=1.5 \times 10^{-2} \mathrm{Nm}
\)

Hint

(c) Force due to electric field acts along the direction of the electric field but force due to the magnetic field acts along a direction perpendicular to both the velocity of the charged particle and the magnetic field. Hence both statements (2) and (3) are true. In statement (2), magnetic force is zero, so, electric force will keep the particle continue to move in horizontal direction. In statement (3), both electric and magnetic forces will be opposite to each other. If their magnitudes will be equal then the particle will continue horizontal motion.

Hint

(b) Force on a charged particle is given by \(F\) \(=q v B\). Here \(v=0\) and also resultant \(B\) is zero.
\(\therefore\) Force \(=0\)

Hint

(b) For the circular motion in a cyclotron,
\(
q v B=\frac{m v^2}{r} \Rightarrow q B=m \omega=\frac{m \times 2 \pi}{T}
\)
\(\therefore \quad T=\frac{2 \pi m}{q B}\) is independent of \(v\) and \(r\).

Hint

(d) The magnetic force acting on the charged paraticle is given by
\(
\begin{aligned}
& \overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}}) \\
& \left.=\left(-2 \times 10^{-6}\right)\left[\{2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \times 10^6\right\} \times(2 \hat{\mathrm{j}})\right] \\
& =-4(2 \hat{\mathrm{k}}) \\
& =-8 \hat{\mathrm{k}}
\end{aligned}
\)
\(\therefore\) Force is of 8 N along negative z -axis.

Hint

(c)

\(
\mathrm{G}=60 \Omega, {I}_{{g}}=1.0 \mathrm{~A}, {I}=5 \mathrm{~A} .
\)
Let \(S\) be the shunt resistance connected in parallel to galvanometer
\(
\begin{aligned}
& {I}_{{g}} \mathrm{G}=\left(I-{I}_{{g}}\right) {S} \\
& S=\frac{I_g G}{I-I_g}=\frac{1}{5-1} \times 60=\frac{1.0 \times 60}{4}=15 \Omega
\end{aligned}
\)
Thus by putting \(15 \Omega\) in parallel, the galvanometer can be converted into an ammeter.

Hint

(b) According to the figure the magnitude of force on the segment \(Q M\) is \(F_3-F_1\) and PM is \({F}_2\).

Therefore, the magnitude of the force on segment PQ is \(\sqrt{\left(F_3-F_1\right)^2+F_2^2}\)

Hint

(a) When a charged particle having a given K.E, \(T\) enters in a field of magnetic induction, which is perpendicular to its velocity, it takes a circular trajectory. It does not increase in energy, therefore \(T\) is the K.E.

Hint

(d)

Current through the galvanometer
\(
I=\frac{3}{(50+2950)}=(10)^{-3} \mathrm{~A}
\)
Current for 30 divisions \(=10^{-3} \mathrm{~A}\)
Current for 20 divisions
\(
=\frac{10^{-3}}{30} \times 20=\frac{2}{3} \times 10^{-3} \mathrm{~A}
\)
To obtain a deflection of 20 divisions, let resistance added be \(R\)
\(
\begin{aligned}
& \therefore \frac{2}{3} \times 10^{-3}=\frac{3}{(50+R)} \\
& \text { or } R=4450 \Omega
\end{aligned}
\)

Hint

(a)

Let \(i_a\) is the current flowing through ammeter and \(i\) is the total current. So, a current \(i-i_a\) will flow through shunt resistance.
Potential difference across ammeter and shunt resistance is same.
ie, \(i_a \times R=\left(i-i_a\right) \times S\)
or \(S=\frac{i_o R}{i-i_a} \dots(i)\)
Given, \(i_a=100 \mathrm{~A}, i=750 \mathrm{~A}, R=13 \Omega\)
Hence, \(S=\frac{100 \times 13}{750-100}=2 \Omega\)

Hint

(a) As revolving charge is equivalent to a current, so
\(
I=q f=q \times \frac{\omega}{2 \pi}
\)
But \(\omega=\frac{v}{R}\)
where \(R\) is radius of circle and \(v\) is uniform speed of charged particle.
Therefore, \(I=\frac{q v}{2 \pi R}\)
Now, magnetic moment associated with charged particle is given by
\(
\begin{aligned}
& \mu=I A=I \times \pi R^2 \\
& \text { or } \mu=\frac{q v}{2 \pi R} \times \pi R^2 \\
& =\frac{1}{2} q v R
\end{aligned}
\)

Hint

(b) The radius of the orbit in which ions moving \(R=\frac{m v}{q B}\)
The angluar frequency of rotation of the ions about the vertical field \(B\) is given by
\(
\omega=\frac{v}{R}=\frac{q B}{m}=2 \pi v
\)
where \(v\) is frerquency.
Energy of ion is given by
\(
E=\frac{1}{2} m v^2=\frac{1}{2} m(R \omega)^2=\frac{1}{2} m R^2 B^2 \frac{q^2}{m^2}
\)
or \(E=\frac{1}{2} \frac{R^2 B^2 q^2}{m} \dots(i)\)
If ions are accelerated by electric potental \(V\), then energy attained by ions
\(
E=q V \dots(ii)
\)
From Eqs. (i) and (ii) and get
\(
\begin{aligned}
& q V=\frac{1}{2} \frac{R^2 B^2 q^2}{m} \\
& \text { or } \frac{q}{m}=\frac{2 V}{R^2 B^2}
\end{aligned}
\)
If \(V\) and \(B\) are kept constant, then \(\frac{q}{m} \propto \frac{1}{R^2}\)

Hint

(b) If both electric and magnetic fields are present and perpendicular to each other and the particle is moving perpendicular to both of

them with \(F_e=F_m\). In this situation \(\vec{E} \neq 0\) and \(\vec{B} \neq 0\)
But if electric field becomes zero, then only force due to magnetic field axists. Under this force, the charge moves along a circle.

Note: If the electric field is switched off, and the same magnetic field is maintained, the electrons move in a circular orbit and electron will travel a magnetic field perpendicular to its velocity.

Hint

(c) Force acting on a charged particle moving with velocity \(\vec{v}\) is subjected to magnetic field \(\vec{B}\) is given by
\(
\vec{F}=q(\vec{v} \times \vec{B}) \quad \text { or } \quad \quad F=q v B \sin \theta
\)
(i) When \(\theta=0^{\circ}, F=q v B \sin 0^{\circ}=0\)
(ii) When \(\theta=90^{\circ}, F=q v B \sin 90^{\circ}=q v B\)
(iii) When \(\theta=180^{\circ}, F=q v B \sin 180^{\circ}=0\)
This implies force acting on a charged particle is nonzero, when angle between \(\vec{v}\) and \(\vec{B}\) can have any value other than zero and \(180^{\circ}\).

Hint

(b) 1. Define the Radii:
Let the radius of coil 2 be \(r\). Then, the radius of coil 1 will be \(r_1=2 r\).
2. Magnetic Field Formula:
The magnetic field \(B\) at the center of a circular coil carrying current \(I\) is given by the formula:
\(
B=\frac{\mu_0 I}{2 R}
\)
where \(R\) is the radius of the coil.
3. Magnetic Fields for Both Coils:
For coil 1:
\(
B_1=\frac{\mu_0 I_1}{2 r_1}=\frac{\mu_0 I_1}{4 r}
\)
For coil 2:
\(
B_2=\frac{\mu_0 I_2}{2 r_2}=\frac{\mu_0 I_2}{2 r}
\)
4. Setting the Magnetic Fields Equal:
Since we want the magnetic fields at the centers of both coils to be equal:
\(
B_1=B_2
\)
This gives us:
\(
\frac{\mu_0 I_1}{4 r}=\frac{\mu_0 I_2}{2 r}
\)
Simplifying this, we can cancel \(\mu_0\) and \(r\) :
\(
\frac{I_1}{4}=\frac{I_2}{2}
\)
Cross-multiplying leads to:
\(
I_1=2 I_2
\)
5. Resistance of the Coils:
The resistance \(R\) of a wire is given by:
\(
R=\frac{\rho L}{A}
\)
Since both coils are made from the same wire, the total length of the wire is the same. The length of wire for coil 1 is \(L_1=2 \pi r_1=4 \pi r\) and for coil 2 is
\(
L_2=2 \pi r_2=2 \pi r
\)
6. Resistance of Each Coil:
The resistance for coil 1 :
\(
R_1=\frac{\rho(4 \pi r)}{A}
\)
The resistance for coil 2 :
\(
R_2=\frac{\rho(2 \pi r)}{A}
\)
7. Applying Ohm’s Law:
The potential difference \(V\) across a coil is given by Ohm’s law:
\(
V=I R
\)
For coill:
\(
V_1=I_1 R_1=I_1\left(\frac{\rho(4 \pi r)}{A}\right)
\)
For coil 2:
\(
V_2=I_2 R_2=I_2\left(\frac{\rho(2 \pi r)}{A}\right)
\)
8. Finding the Ratio of Potential Differences:
Now, substituting \(I_1=2 I_2\) into the equations for \(V_1\) and \(V_2\) :
\(
\begin{aligned}
& V_1=\left(2 I_2\right)\left(\frac{\rho(4 \pi r)}{A}\right)=\frac{8 \pi \rho r I_2}{A} \\
& V_2=I_2\left(\frac{\rho(2 \pi r)}{A}\right)=\frac{2 \pi \rho r I_2}{A}
\end{aligned}
\)
Now, taking the ratio:
\(
\frac{V_1}{V_2}=\frac{\frac{8 \pi \rho I_2}{A}}{\frac{2 \pi \rho \rho I_2}{A}}=\frac{8}{2}=4
\)
Final Answer:
The ratio of potential differences applied across the two coils is:
\(
\frac{V_1}{V_2}=4: 1
\)

Hint

(c) The current flowing clockwise in an equilateral triangle has a magnetic field in the direction of \(\hat{k}\).
\(
\begin{aligned}
& \tau=B i N A \sin \theta \\
& \tau=B i N A \sin 90^{\circ}
\end{aligned}
\)
Area of equilateral triangle \(A=\frac{\sqrt{3}}{4} i^2, N=1\)
\(
\begin{aligned}
& \tau=B i\left(\frac{\sqrt{3}}{4} i^2\right) \\
& i^2=\left(\frac{4 \tau}{\sqrt{3 B i}}\right) \\
& i=\left(\frac{4 \tau}{\sqrt{3 B i}}\right)^{\frac{1}{2}}=2\left(\frac{\tau}{\sqrt{3 B i}}\right)^{1 / 2}
\end{aligned}
\)

Hint

(c) According to Fleming’s left hand rule direction of force is along \(O y\) axis which is perpendicular to wire.

\(
\vec{F}=e(\vec{v} \times \vec{B}) .
\)
\(B\) due to \(i\) is acting inwards \(i . e.\), into the paper. \(v\) is along \(O x\).
\(
\therefore \quad F=Q^{+}\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
v & 0 & 0 \\
0 & 0 & -B
\end{array}\right| \Rightarrow F=Q^{+}[-\hat{j}(-v B)+0]
\)
\(
\therefore \quad \vec{F}=+Q v B \hat{j} . \text { i.e. in } O y \text { direction. }
\)

Hint

(a)

\(
\begin{aligned}
&\text { The magnetic field produce by moving electron in circular path }\\
&\begin{aligned}
& \mathrm{B}=\frac{\mu_0 i}{2 r} \\
& \text { and } {i}=\frac{q}{2 \pi r} \times v \\
& \therefore \mathrm{~B}=\frac{\mu_0 q v}{4 \pi r^2} \\
& \Rightarrow \mathrm{r} \propto \sqrt{\frac{v}{B}}
\end{aligned}
\end{aligned}
\)

Hint

(c) The total current shown by the galvanometer
\(
\begin{aligned}
& =25 \times 4 \times 10^{-4} \mathrm{~A} \\
& I_g=10^{-2} \mathrm{~A}
\end{aligned}
\)
The value of resistance connected in series to convert galvanometer into voltmeter of 25 V is
\(
R=\frac{V}{I_g}-G=\frac{25}{10^{-2}}-50=2450 \Omega
\)

Hint

(a) The resistance of coil of galvanometer of its own is low, hence to convert a galvanometer into a voltmeter, its resistance is to be increased. For this an appropriate high resistance is joined in sereis with the galvanometre as shown in the figure. The value of this resistance to be connected depends on the range of the voltmeter.

Hint

(a) For converting galvanometer to voltmeter, a high resistance should be connected in series with galvanometer.

Hint

(a) If a moving charged particle is subjected to a perpendicular uniform magnetic field, then according to \(F=q v B \sin \theta\), it will experience a maximum force which will provide the centripetal force to particle and it will describe a circular path with uniform speed.

Note: Magnetic force acts perpendicular to the velocity. Hence speed remains constant.

Hint

(b) Magnetic field induction atporint inside the solenoid of length \(l\), having \(n\) türns per unit length carrying current \(i\) is given by
\(
B=\mu_0 n i
\)
If \(i \rightarrow\) doubled \(_s n \rightarrow\) halved then \(B \rightarrow\) remains same.
\(
B_1=\left(\mu_0\right)\left(\frac{n}{2}\right)(2 i)=\mu_0 n i=B
\)

Hint

(b) The force experienced by a charged particle moving in space where electric and magnetic field exists is called Lorentz force.

\(
\begin{aligned}
&\text { Lorentz force }\\
&\begin{aligned}
& \overrightarrow{F_L}=\overrightarrow{F_e}+\overrightarrow{F_m} \\
& =q \vec{E}+q(\vec{v} \times \vec{B})
\end{aligned}
\end{aligned}
\)

Alternate:

Force due to electric field \(=q \vec{E}\)

Force due to magnetic field \(=q(\vec{v} \times \vec{B})\)
Net force experienced \(=q \vec{E}+q(\vec{v} \times \stackrel{\rightharpoonup}{B})\)

Hint

(c) Let \(I\) be current and \(l\) be the length of the wire.
For Ist case : \(B=\frac{\mu_0 I n}{2 r}=\frac{\mu_0 I \times \pi}{l}\) where \(2 \pi r=l\) and \(n=1\)
For IInd Case : \(l=2\left(2 \pi r^{\prime}\right) \Rightarrow r^{\prime}=\frac{l}{4 \pi}\)
\(
B^{\prime}=\frac{\mu_0 n I}{2 r^{\prime}}=\frac{\mu_0 2 I}{2 \frac{l}{4 \pi}}=\frac{4 \mu_0 I \pi}{l}=4 B
\)

Hint

\(
\text { (a) Magnetic moment } M=n i \mathrm{~A}
\)

Hint

(b) A tangent galvanometer is an instrument used to measure the current in a circuit. It works on the principle that the deflection of a magnetic needle is proportional to the current passing through a coil in a magnetic field. The deflection is measured in terms of the angle, and the current is determined using the tangent of that angle.

Explanation:
The galvanometer consists of a coil of wire mounted on a horizontal axis, with a compass needle at the center. When a current passes through the coil, it generates a magnetic field that interacts with the needle, causing it to deflect.
The amount of deflection \((\theta)\) is related to the current \((I)\) passing through the coil by the equation:
\(
\tan (\theta)=\frac{k I}{H}
\)
where:
\(\theta\) is the angle of deflection,
\(\boldsymbol{k}\) is a constant that depends on the geometry of the galvanometer,
\(H\) is the horizontal component of the Earth’s magnetic field,
\(I\) is the current to be measured.

Hint

(b) Electron moves undeflected if force exerted due to electric field is equal to force due to magnetic field.
\(
q|\vec{v} \| \vec{B}|=q|\vec{E}| \Rightarrow|\vec{v}|=\frac{|\vec{E}|}{|\vec{B}|}
\)

Hint

(c) The frequency of revolution of charged particle in a perpendicular magnetic field is
\(
\begin{aligned}
& f=\frac{1}{T}=\frac{1}{\frac{2 \pi r}{v}}=\frac{v}{2 \pi r} \\
& =\frac{v}{2 \pi} \times \frac{e B}{m v}=\frac{e B}{2 \pi m}
\end{aligned}
\)

Hint

\(
\text { (c) } \begin{aligned}
& B=\frac{\mu_0}{4 \pi} \frac{2 i_2}{(r / 2)}-\frac{\mu_0}{4 \pi(r / 2)} \frac{2 i_1}{(r / 2}=\frac{\mu_0}{4 \pi} \frac{4}{r}\left(i_2-i_1\right) \\
& =\frac{\mu_0}{4 \pi} \frac{4}{5}(2.5-5.0)=-\frac{\mu_0}{2 \pi} .
\end{aligned}
\)
-ve sign show that \(B\) is acting inwards i.e, into the plane.

Hint

\(
\begin{aligned}
& \text { (b) } r=\frac{m v \sin \theta}{B e}=\frac{3 \times 10^5 \sin 30^{\circ}}{0.3 \times 10^8} \\
& \frac{3 \times 10^5 \times \frac{1}{2}}{3 \times 10^7}=0.5 \times 10^{-2} \mathrm{~m}=0.5 \mathrm{~cm} .
\end{aligned}
\)

Note: If the change particle is moving perpendicular to the magnetic field then \(q V B=\frac{m v^2}{r}\)
\(
\Rightarrow r=\frac{m v}{q B}
\)
If the charge particle is moving at an, angle (other than \(0^{\circ}, 90^{\circ}, 180^{\circ}\) ) to the field, then
\(
q(v \sin \theta) B=\frac{m(v \sin \theta)^2}{r} \Rightarrow r=\frac{m v \sin \theta}{q B}
\)

Hint

(a) Given : Two charged particles \(A\) & \(B\) at rest ; potential difference \(=V\);
\(
\text { magnetic field }=B ; \text { radii }=r_1 \text { and } r_2
\)
Let radius of A and b be \(r_1\) and \(r_2\) and
their masses \(m_1\) and \(m_2\)
\(
r=\frac{\sqrt{2 m(q v)}}{q B}
\)
Since they have same charge
\(
\begin{aligned}
& \frac{r_1}{r_2}=\sqrt{\frac{m_1}{m_2}} \\
& \frac{m_1}{m_2}=\left(\frac{r_1}{r_2}\right)^2
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
&\text { Magnetic field at the centre of coil } B=\frac{\mu_0 i N}{2 a}\\
&\begin{aligned}
& \frac{4 \pi \times 10^{-7} \times 5 \times 50}{2 \times 10 / 100}=1.57 \times 10^{-3} T \\
& =1.57 \mathrm{mT}
\end{aligned}
\end{aligned}
\)

Hint

(c) We know that magnetic field at the centre of circular coil,
\(
B=\frac{\mu_0 I n}{2 r}=\frac{4 \pi \times 10^{-7} \times 2 \times 50}{2 \times 0.5}=1.25 \times 10^{-4} \mathrm{~N}
\)

Hint

(b) Inside a hollow pipe carrying current, the magnetic field is zero, since according to Ampere’s law, \(B_i \cdot 2 \pi r=\mu_0 \times 0 \Rightarrow B_i=0\).
But for external points, the current behaves as if it was concentrated at the axis only; so, outside, \(B_0=\frac{\mu_0 i}{2 \pi r}\). Thus, the magnetic field is produced outside the pipe only.

Hint

(b) Diameter of first wire \(\left(d_1\right)=0.5 \mathrm{~mm}\); Current in first wire \(\left(I_1\right)=1 \mathrm{~A}\); Diameter of second wire \(\left(d_2\right)=1 \mathrm{~mm}\) and current in second wire \(\left(I_2\right)=1 \mathrm{~A}\). Strength of magnetic field due to current flowing in a conductor, \((B)=\frac{\mu_0}{4 \pi} \times \frac{2 I}{a}\) or \(B \propto I\).
Since the current in both the wires is same, therefore there is no change in the strength of the magnetic field.

Hint

(c) Potential difference \((\mathrm{V})=1 \mathrm{~V}\),
K.E. acquired \(=q V\)
\(
\begin{aligned}
& =1.6 \times 10^{-19} \times 1 \\
& =1.6 \times 10^{19} \text { joules }=1 \mathrm{eV}
\end{aligned}
\)

Hint

(a)

\(
B=\frac{\mu_0 N i}{2 r}=\frac{4 \pi \times 10^{-7} \times 1000 \times 0.1}{2 \times 0.1}=6.28 \times 10^{-4} \mathrm{~T}
\)

Hint

\(
\begin{aligned}
& \text { (d) }: q v B \sin \theta=\frac{m v^2}{R} \\
& R=\frac{m v}{q B \sin \theta}=\frac{3 \times 10^5}{10^8 \times 0.3 \times \frac{1}{2}}=2 \mathrm{~cm}
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
F & =\frac{\mu_0}{4 \pi} \frac{2 I_1 I_2}{r} \\
& =\frac{10^{-7} \times 2(1) \times(1)}{1}=2 \times 10^{-7} \mathrm{~N} / \mathrm{m}
\end{aligned}
\)

Hint

(b) \(\text { Here, } G=8 \Omega, S=2 \Omega \text { and } I=1 A \text {. }\)

Current through shunt, \(I_s=I \frac{G}{(G+S)}\) \(=1 \times \frac{8}{(8+2)}=0.8 \mathrm{~A}\)

Hint

(b) Let \(\ell\) be length of wire.
Ist case \(: \ell=2 \pi r \Rightarrow r=\frac{\ell}{2 \pi}\)
\(
B=\frac{\mu_0 I n}{2 r}=\frac{\mu_0 I \times 2 \pi}{2 \ell}=\frac{\mu_0 \pi I}{\ell}[\because n=1] \ldots(1)
\)
2nd Case : \(\ell=2\left(2 \pi r^{\prime}\right) \Rightarrow r^{\prime}=\frac{\ell}{4 \pi}\)
\(
B^{\prime}=\frac{\mu_0 I n}{2 \frac{\ell}{4 \pi}}=\frac{2 \mu_0 I \pi}{\frac{\ell}{2}}=4\left(\frac{\mu_0 \pi I}{\ell}\right)=4 B,
\)
using (1) (where \(n=2\) )
\(
\text { Therefore, } \frac{B}{B^{\prime}}=\frac{1}{4}
\)

Hint

(c) Distance between two parallel wires \((x)=10 \mathrm{cm}=0.1 \mathrm{~m}\),
Current in each wire \(=I_1=I_2=10 \mathrm{~A}\) and length of wire \((l)=1 \mathrm{~m}\).
Force on the wire \((F)=\frac{\mu_0 I_1 \cdot I_2 \times l}{2 \pi x}\)
\(
=\frac{\left(4 \pi \times 10^{-7}\right) \times 10 \times 10 \times 1}{2 \pi \times 0.1}=2 \times 10^{-4} \mathrm{~N} .
\)
Since the current is flowing in the same direction, therefore the force will be attractive.

Hint

(b) When a positively charged particle enters in a region of uniform magnetic field, directed vertically upwards, it experiences a centripetal force which will move it in circular path with a uniform speed.

Hint

(b) Tesla is the unit of magnetic field.

Note: The tesla (symbol: T ) is the unit of magnetic flux density (also called magnetic B-field strength) in the International System of Units (SI). One tesla is equal to one weber per square metre.

Hint

(b) Net magnetic field on AB is zero because magnetic field due to both current carrying wires is equal in magnitude but opposite in direction.

Note: The direction of the magnetic field, due to current, is given by the right-hand rule. At axis AB, the components of magnetic field will cancel each other and the resultant magnetic field will be zero.

Hint

(d)

\(
\begin{aligned}
&\text { According to Biot Savart law, }\\
&d \vec{B}=\frac{\mu_0}{4 \pi} \frac{i(d \vec{\ell} \times \vec{r})}{r^3}
\end{aligned}
\)

Hint

(a) K.E. of electron \(=10 \mathrm{eV}\)
\(
\begin{aligned}
& \Rightarrow \frac{1}{2} m v^2=10 \mathrm{eV} \\
& \Rightarrow \frac{1}{2}\left(9.1 \times 10^{-31}\right) v^2=10 \times 1.6 \times 10^{-19} \\
& \Rightarrow v^2=\frac{2 \times 10 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}} \\
& \Rightarrow v^2=3.52 \times 10^{12} \Rightarrow v=1.88 \times 10^6 \mathrm{~m}
\end{aligned}
\)
Also, we know that for circular motion
\(
\frac{m v^2}{r}=B e v \Rightarrow r=\frac{m v}{B e}=11 \mathrm{~cm}
\)

Hint

(c) The electron moves with constant velocity without deflection. Hence, force due to magnetic field is equal and opposite to force due to electric field.
\(
q v B=q E \Rightarrow v=\frac{E}{B}=\frac{20}{0.5}=40 \mathrm{~m} / \mathrm{s}
\)

Hint

(a) Current \((\mathrm{I})=12 \mathrm{~A}\) and magnetic field (B) \(=3 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^2\). Consider magnetic field \(\vec{B}\) at distance \(r\).
Magnetic field, \(B=\frac{\mu_0 I}{2 \pi r}\)
\(
\Rightarrow r=\frac{\mu_0 I}{2 \pi B}=\frac{\left(4 \pi \times 10^{-7}\right) \times 12}{2 \times \pi \times\left(3 \times 10^{-5}\right)}=8 \times 10^{-2} \mathrm{~m}
\)

Hint

(d) When the deflection produced by electric field is equal to the deflection produced by magnetic field, then the electron can go undeflected.

Hint

(c) Area \((A)=0.01 \mathrm{~m}^2\); Current \((I)=10 \mathrm{~A}\);
Angle \((\phi)=90^{\circ}\) and magnetic field \((B)=0.1 \mathrm{~T}\).
Therefore acutal angle \(\theta=\left(90^{\circ}-\phi\right)=\left(90^{\circ}-90^{\circ}\right)=0^{\circ}\).
And torque acting on the loop \((\tau)=I A B \sin \theta\)
\(
=10 \times 0.01 \times 0.1 \times \sin 0^{\circ}=0
\)

Hint

(a) The force acting on a charged particle in magnetic field is given by
\(
\vec{F}=q(\vec{v} \times \vec{B}) \text { or } F=q v B \sin \theta,
\)
When angle between v and B is \(180^{\circ}\),
\(
F=0
\)
Force, \(\vec{F}=\vec{q}(\vec{v} \times \vec{B})=q v B \sin \theta\)
When \(\theta=0^{\circ}, 180^{\circ}\) then particle will be moving in a direction parallel or antiparallel to the field. In such cases, the trajectory of the particle is a straight line.

Hint

(a) A current carrying coil has magnetic dipole moment. Hence a torque \(p_m \times B\) acts on it in magnetic field.

Hint

(a) To convert a galvanometer into an ammeter, one needs to connect a low resistance in parallel so that maximum current passes through the shunt wire and ammeter remains protected.

Hint

\(
\text { (b)  } F=i l \times B=1.2 \times 0.5 \times 2=1.2 \mathrm{~N} \text {. }
\)

Hint

(a) \(B=\frac{\mu_0 i}{2 \pi r}\) or \(B \propto \frac{1}{r}\)

When \(r\) is doubled, the magnetic field becomes halved \(i. e\), now the magnetic field will be 0.2 T.

Hint

(c) \(r=\frac{m v}{q B}\) or \(r \propto v\)

As \(v\) is doubled, the radius also becomes double.
Hence, radius \(=2 \times 2=4 \mathrm{~cm}\)

Hint

(d) For a charged particle orbiting in a circular path in a magnetic field
\(
\frac{m v^2}{r}=B q v \Rightarrow v=\frac{B q r}{m}
\)
or, \(m v^2=B q v r\)
Also,
\(
E_K=\frac{1}{2} m v^2=\frac{1}{2} B q v r=B q \frac{r}{2} \cdot \frac{B q r}{m}=\frac{B^2 q^2 r^2}{2 m}
\)
For deuteron, \(E_1=\frac{B^2 q^2 r^2}{2 \times 2 m}\)
For proton, \(E_2=\frac{B^2 q^2 r^2}{2 m}\)
\(
\frac{E_1}{E_2}=\frac{1}{2} \Rightarrow \frac{50 \mathrm{keV}}{E_2}=\frac{1}{2} \Rightarrow E_2=100 \mathrm{keV}
\)

Hint

(a) \(B=\frac{\mu_0 I}{2 \pi r} \Rightarrow B \propto \frac{1}{r}\)

As the distance is increased to three times, the magnetic induction reduces to one third. Hence,
\(
B=\frac{1}{3} \times 10^{-3} \text { tesla }=3.33 \times 10^{-4} \text { tesla }
\)

Hint

(b) Energy is stored in magnetic field.

Hint

(d) The plane of coil will orient itself so that area vector aligns itself along the magnetic field.
So, the plane will orient perpendicular to the magnetic field.