Past NEET Entrance Papers

Hint

(b)

(A) hold the sheet there if the metal is magnetic
(C) move the sheet away from the pole with uniform velocity if the metal is magnetic
(D) move the sheet away from the pole with uniform velocity if the metal is nonmagnetic

The strong magnetic pole will attract the magnet, so a force is needed to hold the sheet there if the metal is magnetic.

If we move the metal sheet (magnetic or nonmagnetic) away from the pole, eddy currents are induced in the sheet. Because of eddy currents, thermal energy is produced in it. This energy comes at the cost of the kinetic energy of the plate; thus, the plate slows down. So, a force is needed to move the sheet away from the pole with uniform velocity.

Hint

(a) We are given the equation for the current in an inductor-resistor-battery circuit:
\(
i=i_0\left(1-e^{-\frac{t}{\tau}}\right)
\)
Substitute \(i=0.5 i_0\) into the equation for \(i\) :
\(
0.5 i_0=i_0\left(1-e^{-\frac{t}{\tau}}\right)
\)
\(
0.5=1-e^{-\frac{t}{t}}
\)
\(
e^{-\frac{t}{\tau}}=1-0.5=0.5
\)
\(
-\frac{t}{\tau}=\ln (0.5)
\)
\(
\begin{aligned}
&\text { Using the property } \ln (0.5)=-\ln (2) \text {, and given that } \ln (2)=0.693 \text {, we get: }\\
&-\frac{t}{\tau}=-0.693
\end{aligned}
\)
\(
t=0.693 \tau [\tau=L/R=10 \times 10^{-3} s]
\)
\(
t=6.93 ms
\)

Hint

(b) The flux through the loop is:
\(
\Phi_B(t)=B(t) \cdot A=(0.5 \sin (100 \pi t)) \cdot\left(2.5 \times 10^{-3}\right)=1.25 \times 10^{-3} \sin (100 \pi t) \mathrm{Wb} .
\)
Now, we differentiate the flux with respect to time to get the induced emf:
\(
\mathcal{E}=-\frac{d \Phi_B}{d t}=-\frac{d}{d t}\left(1.25 \times 10^{-3} \sin (100 \pi t)\right)
\)
We differentiate \(\sin (100 \pi t)\) using the chain rule:
\(
\frac{d}{d t}(\sin (100 \pi t))=100 \pi \cos (100 \pi t)
\)
Thus:
\(
\mathcal{E}(t)=-1.25 \times 10^{-3} \cdot 100 \pi \cos (100 \pi t)=-1.25 \times 10^{-1} \pi \cos (100 \pi t) \mathrm{V}
\)
\(
\begin{aligned}
&\text { At } t=0, \cos (100 \pi t)=\cos (0)=1 \text {, so: }\\
&\mathcal{E}(0)=-1.25 \times 10^{-1} \pi \cdot 1=-1.25 \times 10^{-1} \pi \mathrm{~V}=
\end{aligned}
\)
\(
|\mathcal{E}(0)|=0.125 \pi \mathrm{~V}=125 \pi \mathrm{mV}
\)

Hint

(a) As the rod rotates, each point on the rod moves in a circular path. The velocity of a point at a distance \(r\) from the center of the rod is given by:
\(
v(r)=\omega r
\)
Where:
\(\omega\) is the angular velocity,
\(r\) is the distance from the center of the rod.
Now, when the rod moves in the presence of a magnetic field, the velocity of each point contributes to the generation of an electromotive force (emf) due to the magnetic induction. The induced emf at a distance \(r\) from the center is given by:
\(
d \mathcal{E}=B v(r) d r
\)
Substituting for \(v(r)\) :
\(
d \mathcal{E}=B(\omega r) d r=B \omega r d r
\)
To find the total emf between the center and an end of the rod, we integrate this expression from \(r=0\) to \(r=\frac{L}{2}\), where \(\frac{L}{2}\) is the distance from the center to an end of the rod.
\(
\mathcal{E}=\int_0^{L / 2} B \omega r d r
\)
Perform the integration:
\(
\begin{gathered}
\mathcal{E}=B \omega \int_0^{L / 2} r d r=B \omega\left[\frac{r^2}{2}\right]_0^{L / 2} \\
\mathcal{E}=B \omega\left(\frac{(L / 2)^2}{2}\right) \\
\mathcal{E}=B \omega \cdot \frac{L^2}{8}
\end{gathered}
\)

Hint

(a)
\(
\begin{aligned}
&\text { The self-inductance } L \text { of a solenoid is given by the formula: }\\
&L=\mu_0 \mu_r \frac{N^2 A}{l}
\end{aligned}
\)
\(\mu_0\) is the permeability of free space,
\(\mu_r\) is the relative permeability of the material,
\(N\) is the total number of turns in the solenoid,
\(A\) is the cross-sectional area of the solenoid, \(l\) is the length of the solenoid.
The total number of turns \(N\) in each solenoid is related to the number of turns per unit length and the length of the solenoid:
For solenoid \(A\), the total number of turns is:
\(
N_A=\left(\frac{N_A}{L_A}\right) L_A=\left(\frac{1}{2} \cdot \frac{N_B}{L_B}\right) \cdot 2 L_B=N_B
\)
Thus, \(N_A=N_B\), i.e., both solenoids have the same total number of turns.
For solenoid \(A\) :
\(
L_A=\mu_0 \mu_r \frac{N_A^2 A}{L_A}
\)
For solenoid \(B\) :
\(
L_B=\mu_0 \mu_r \frac{N_B^2 A}{L_B}
\)
Taking the Ratio \(L_A: L_B\)
We now take the ratio of the inductances \(L_A\) and \(L_B\) :
\(
\frac{L_A}{L_B}=\frac{\mu_0 \mu_r \frac{N_A^2 A}{L_A}}{\mu_0 \mu_r \frac{N_B^2 A}{L_B}}=\frac{N_A^2}{N_B^2} \cdot \frac{L_B}{L_A}
\)
Since \(N_A=N_B\), the ratio simplifies to:
\(
\frac{L_A}{L_B}=\frac{L_B}{L_A}
\)
Substituting \(L_A=2 L_B\) :
\(
\frac{L_A}{L_B}=\frac{L_B}{2 L_B}=\frac{1}{2}
\)
Thus, the ratio of the self-inductances of the two solenoids is:
\(
L_A: L_B=1: 2
\)

Hint

(a) The magnetic energy stored in an inductor is given by the formula:
\(
U=\frac{1}{2} L I^2
\)
Where:
\(U\) is the magnetic energy stored in the inductor,
\(L\) is the inductance of the inductor,
\(I\) is the current flowing through the inductor.

Given:
The inductance \(L=4 \mu \mathrm{H}=4 \times 10^{-6} \mathrm{H}_{\text {, }}\)
The current \(I=2 \mathrm{~A}\).
Substituting the values into the formula:
\(
\begin{gathered}
U=\frac{1}{2} \times\left(4 \times 10^{-6}\right) \times(2)^2 \\
U=\frac{1}{2} \times 4 \times 10^{-6} \times 4 \\
U=8 \times 10^{-6} \mathrm{~J}=8 \mu \mathrm{~J}
\end{gathered}
\)

Hint

(a) The maximum induced electromotive force (emf) in an AC generator is given by the formula:
\(
\mathcal{E}_{\max }=N B A \omega
\)
Where:
\(\mathcal{E}_{\text {max }}\) is the maximum emf,
\(N\) is the number of turns of the coil,
\(B\) is the magnetic field strength,
\(A\) is the area of the coil,
\(\omega\) is the angular velocity of the coil.
Given:
Number of turns, \(N=100\),
Magnetic field, \(B=0.05 \mathrm{~T}\),
Area of the coil, \(A=1 \mathrm{~m}^2\),
The coil makes one revolution per second, so the angular velocity \(\omega=2 \pi \mathrm{rad} / \mathrm{s}\) (since one revolution corresponds to \(2 \pi\) radians).
Substituting the values into the formula:
\(
\begin{gathered}
\mathcal{E}_{\max }=100 \times 0.05 \times 1 \times 2 \pi \\
\mathcal{E}_{\max }=100 \times 0.05 \times 2 \pi \\
\mathcal{E}_{\max }=10 \pi \mathrm{~V} \\
\mathcal{E}_{\max } \approx 31.42 \mathrm{~V}
\end{gathered}
\)

Hint

(d) The net magnetic flux through any closed surface is zero.
This result comes from Gauss’s law for magnetism, which states that:
\(
\oint_{\text {closed surface }} \mathbf{B} \cdot d \mathbf{A}=0
\)
This implies that the total magnetic flux through any closed surface is zero because there are no “magnetic charges” (monopoles). Magnetic field lines always form closed loops, meaning they enter and exit a surface in equal amounts, canceling each other out.

Hint

(c) The magnetic flux \(\Phi_B\) through a surface is given by the formula:
\(
\Phi_B=B \cdot A \cdot \cos \theta
\)
Where:
\(B\) is the magnetic field strength,
\(A\) is the area of the loop,
\(\theta\) is the angle between the magnetic field and the normal to the surface of the loop.
Given:
Side of the square loop, \(s=1 \mathrm{~m}\),
Resistance of the loop, \(R=1 \Omega\) (not directly relevant for this part of the problem),
Magnetic field strength, \(B=0.5 \mathrm{~T}\),
The plane of the loop is perpendicular to the magnetic field, so \(\theta=0^{\circ}\).
Step 1: Calculate the area of the square loop:
\(
A=s^2=1^2=1 \mathrm{~m}^2
\)
Step 2: Calculate the magnetic flux:
Since \(\theta=0^{\circ}\) (the plane of the loop is perpendicular to the magnetic field, meaning the magnetic field is normal to the loop), \(\cos 0^{\circ}=1\). Therefore, the flux is:
\(
\Phi_B=B \cdot A \cdot \cos 0^{\circ}=\downarrow \mathrm{T} \times 1 \mathrm{~m}^2 \times 1=0.5 \text { weber }
\)

Hint

(d) The induced emf \(\mathcal{E}\) in a rotating coil is given by:
\(
\mathcal{E}=N B A \omega \cos (\theta)
\)
Where:
\(N\) is the number of turns,
\(B\) is the magnetic field strength,
\(A\) is the area of the coil,
\(\omega\) is the angular velocity,
\(\theta\) is the angle between the magnetic field and the normal to the plane of the coil.
Since the coil is rotating about its horizontal diameter, the angle between the magnetic field and the plane of the coil changes with time. The maximיm induced emf occurs when \(\cos (\theta)=1\), which happens when the plane of the coil is perpendicular to the magnetic field.
The area \(\boldsymbol{A}\) of a circular coil is:
\(
A=\pi r^2
\)
Substituting \(r=10 \mathrm{~m}\) :
\(
A=\pi \times(10)^2=100 \pi \mathrm{~m}^2
\)
\(
\begin{gathered}
\mathcal{E}_{\max }=N B A \omega \\
\mathcal{E}_{\max }=1000 \times\left(2 \times 10^{-5}\right) \times 100 \pi \times 2 \\
\mathcal{E}_{\max }=1000 \times 2 \times 10^{-5} \times 100 \times \pi \times 2 \\
\mathcal{E}_{\max }=4 \pi \mathrm{~V}
\end{gathered}
\)
Using Ohm’s law, the induced current is given by:
\(
I=\frac{\mathcal{E}}{R}
\)
Substituting the values:
\(
\begin{gathered}
I=\frac{12.56}{12.56} \\
I \approx 1 \mathrm{~A}
\end{gathered}
\)

Hint

(d) To calculate the induced emf in the coil, we use Faraday’s law of induction, which states that the induced emf \(\mathcal{E}\) in an inductor is related to the rate of change of current through it:
\(
\mathcal{E}=-L \frac{d i}{d t}
\)
Where:
\(L\) is the self-inductance of the coil,
\(\frac{d i}{d t}\) is the rate of change of current.
Given:
The self-inductance \(L=4 \mathrm{H}\),
The current changes from 4 A to 2 A in 1 second, so the change in current \(\Delta i=2 \mathrm{~A}\),
The time interval \(\Delta t=1 \mathrm{~s}\).
Step 1: Calculate the rate of change of current:
\(
\frac{d i}{d t}=\frac{\Delta i}{\Delta t}=\frac{2-4}{1}=-2 \mathrm{~A} / \mathrm{s}
\)
Step 2: Calculate the induced emf:
Now, substituting the values into the formula for the induced emf:
\(
\mathcal{E}=-L \frac{d i}{d t}=-4 \times(-2)=8 \mathrm{~V}
\)

Hint

(a) The mutual inductance \(M\) is a measure of the ability of one coil to induce an emf in another coil due to a changing current. The formula for mutual inductance is:
\(
\mathcal{E}_2=-M \frac{d i_1}{d t}
\)
Where:
\(\mathcal{E}_2\) is the induced emf in the second coil,
\(M\) is the mutual inductance,
\(\frac{d i_1}{d t}\) is the rate of change of current in the first coil.
Step 1: Dimensions of emf
The dimensions of \(\operatorname{emf}(\mathcal{E})\) are the same as the dimensions of voltage, which are:
\(
[\mathcal{E}]=[V]=\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{~A}^{-1}
\)
Step 2: Dimensions of current
The dimensions of current \(i\) are:
\(
[i]=\mathrm{A}
\)
Step 3: Dimensions of mutual inductance
From the formula \(\mathcal{E}_2=-M \frac{d i_1}{d t}\), the dimensir \(\downarrow\) of mutual inductance can be derived by rearranging the formula:
\(
M=\frac{\mathcal{E}_2}{\frac{d i_1}{d t}}
\)
Substituting the dimensions of \(\mathcal{E}_2\) and \(\frac{d i_1}{d t}\) :
\(
[M]=\frac{\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{~A}^{-1}}{\mathrm{~A} \cdot \mathrm{~T}^{-1}}=\mathrm{M} \cdot \mathrm{~L}^2 \cdot \mathrm{~T}^{-2} \cdot \mathrm{~A}^{-2}
\)
Thus, the dimensions of mutual inductance are:
\(
[M]=\mathrm{M} \cdot \mathrm{~L}^2 \cdot \mathrm{~T}^{-2} \cdot \mathrm{~A}^{-2}
\)
Final Answer:
The dimensions of mutual inductance \(M\) are:
\(
[M]=\mathrm{M} \cdot \mathrm{~L}^2 \cdot \mathrm{~T}^{-2} \cdot \mathrm{~A}^{-2}
\)

Hint

(d) 

\(
\begin{aligned}
&\text { The magnetic field energy stored in an inductor is given by the formula: }\\
&U=\frac{1}{2} L I^2
\end{aligned}
\)
Given:
Self-inductance, \(L=10 \mathrm{H}\),
Current, \(I=1 \mathrm{~A}\).
\(
\begin{gathered}
U=\frac{1}{2} \times 10 \times(1)^2 \\
U=\frac{1}{2} \times 10 \times 1=5 \mathrm{~J}
\end{gathered}
\)

Hint

(d) Given magnetic flux is
\(
\phi=2 t^3+4 t^2+2 t+5
\)
Induced emf is given by
\(
\begin{aligned}
& \varepsilon=\frac{-d \phi}{d t}=\frac{-d}{d t}\left(2 t^3+4 t^2+2 t+5\right) \\
& |\varepsilon|=6 t^2+8 t+2
\end{aligned}
\)
At \(t=5 \mathrm{~s}\),
\(
\begin{aligned}
& |\varepsilon|=6 \times 25+8 \times 5+2 \\
& =150+40+2 \\
& =192 \mathrm{~V}
\end{aligned}
\)

Hint

(d) \(\text { Two concentric coils are of radius } R_1 \text { and } R_2 \text { as shown }\)

Let a current \(I_1\) flows through the outer circular coil of radius \(R_1\)
The magnetic field at the centre of the coil is
\(
B_1=\frac{\mu_0 I_1}{2 R_1}
\)
As the inner coil of radius \(R_2\) placed co-axially has small radius ( \(R_2<R_1\) ), therefore \(B_1\) may be taken constant over its crosssectional area
Hence, flux associated with the inner coil is
\(
\phi_2=B_1 \pi R_2^2=\frac{\mu_0 I_1}{2 R_1} \pi R_2^2
\)
As \(M=\frac{\phi_2}{I_1}=\frac{\mu_0 \pi R_2^2}{2 R_1}\)
\(
\therefore M \propto \frac{R_2^2}{R_1}
\)

Hint

(b) Step 1: Differentiate the flux expression with respect to time \(t\) :
\(
\frac{d \phi}{d t}=\frac{d}{d t}\left(5 t^2+3 t+60\right)
\)
Using basic differentiation rules:
\(
\frac{d \phi}{d t}=10 t+3
\)
Step 2: Calculate the induced emf at \(t=4 \mathrm{~s}\) :
Substitute \(t=4\) into the derivative of the flux:
\(
\frac{d \phi}{d t}=10(4)+3=40+3=43 \mathrm{~Wb} / \mathrm{s}
\)
Step 3: Find the induced emf:
\(
\mathcal{E}=-\frac{d \phi}{d t}=-43 \mathrm{~V}
\)
The magnitude of the induced emf is:
\(
|\mathcal{E}|=43 \mathrm{~V}
\)

Hint

(a) Given:
Number of spokes, \(N=20\),
Length of each spoke, \(l=1 \mathrm{~m}\),
Rotational speed, \(\omega=120 \mathrm{rpm}\),
Magnetic field, \(B=0.4 \mathrm{G}=0.4 \times 10^{-4} \mathrm{~T}\) (since \(1 \mathrm{G}=10^{-4} \mathrm{~T}\) ),
The wheel is rotating in a plane perpendicular to the magnetic field.
Step 1: Convert rotational speed to radians per second
The angular speed \(\omega\) in radians per second is:
\(
\omega=120 \mathrm{rpm} \times \frac{2 \pi}{60}=4 \pi \mathrm{~rad} / \mathrm{s}
\)
The induced emf \((\varepsilon)\) can be calculated using the formula:
\(
\epsilon=-\frac{d \Phi}{d t}
\)
Where:
\(\Phi\) is the magnetic flux, given by \(\Phi=B \cdot A\)
A is the area swept by the spoke.
For one spoke, the area (A) swept out in one complete rotation can be calculated as:
\(
A=\frac{1}{2} \cdot r^2 \cdot \theta
\)
Where \(\theta\) is the angle in radians. In one second, the angle subtendedis:
\(
\theta=\omega \cdot t=4 \pi \cdot 1=4 \pi \text { radians }
\)
Thus, the area becomes:
\(
A=\frac{1}{2} \cdot r^2 \cdot 4 \pi=2 \pi r^2
\)
Substituting \(r=1 \mathrm{~m}\) :
\(
A=2 \pi \cdot(1)^2=2 \pi \mathrm{~m}^2
\)
Step 4: Calculate the magnetic flux
Now, substituting the values into the magnetic flux equation:
\(
\Phi=B \cdot A=0.4 \times 10^{-4} \cdot 2 \pi
\)
Step 5: Find the induced emf
Now, substituting this into the induced emf formula:
\(
\epsilon=\frac{1}{2} B \omega r^2
\)
Substituting the values:
\(
\begin{aligned}
& \epsilon=\frac{1}{2} \cdot\left(0.4 \times 10^{-4}\right) \cdot(4 \pi) \cdot(1)^2 \\
& \epsilon=0.2 \times 10^{-4} \cdot 4 \pi=0.8 \pi \times 10^{-4} \mathrm{~V}
\end{aligned}
\)
Step 6: Calculate the numerical value
Using \(\pi \approx 3.14\) :
\(
\epsilon \approx 0.8 \cdot 3.14 \times 10^{-4} \approx 2.512 \times 10^{-4} \mathrm{~V}
\)
Thus, the induced emf is approximately:
\(
\epsilon \approx 2.5 \times 10^{-4} \mathrm{~V}
\)

Hint

(a) Concept:
Eddy current- Eddy current is defined as the induced electromotive force produced in the coil when there is a change of magnetic flux linked to the given coil or we can say that the eddy current is the loops of electrical current induced within the conductors by changing magnetic field in the conductor. It is called as eddy current.

EXPLANATION:
1. Induction furnace- An induction furnace, it is having a coil that is carrying an alternating current that surrounds the chamber of the metal. In this furnace, the eddy currents are induced in the metal and the circulation of these currents will be producing extremely high temperatures which make metal melt.
2. Magnetic braking in trains – In trains, the eddy current is used in the brakes. It is used to slow or stop a moving object by dissipating its kinetic energy.
3. Electromagnet- Electromagnetic is the substance in which a magnetic field is produced by supplying the electric current through the material. When the electric current is supplying through the material, there is a change of magnetic flux into the material which produces loops.
4. Electric heater- The electric heater works on the principle of Joule’s heating effect. It states that the heat produced in the resistor is directly proportional to the resistance for a given current and the time in which current is flowing through the conductor.

Hint

(b) Step 1:
\(
\begin{aligned}
&\text { The magnetic flux } \phi \text { through the coil is given by: }\\
&\phi=B A \cos \theta
\end{aligned}
\)
Step 2: Change in flux
Initially, the coil is placed perpendicular to the magnetic field, so \(\theta=0^{\circ}\), and the flux is:
\(
\phi_{\text {initial }}=B A \cos 0^{\circ}=B A
\)
After rotating the coil by \(90^{\circ}\), the angle between the normal to the coil and the magnetic field becomes \(90^{\circ}\), and the flux is:
\(
\phi_{\mathrm{final}}=B A \cos 90^{\circ}=0
\)
\(
\begin{aligned}
&\text { So, the change in magnetic flux } \Delta \phi \text { is: }\\
&\Delta \phi=\phi_{\text {inal }}-\phi_{\text {initial }}=0-B A=-B A
\end{aligned}
\)
Step 3: Induced emf
The induced emf is the rate of change of magnetic flux:
\(
\mathcal{E}=-N \frac{\Delta \phi}{\Delta t}
\)
Where:
\(N=800\) is the number of turns,
\(\Delta t=0.1 \mathrm{~s}\) is the time taken for the rotation.
Now, substituting the known values:
\(
\mathcal{E}=-800 \times \frac{-B A}{0.1}
\)
Substituting \(B=5 \times 10^{-5} \mathrm{~T}\) and \(A=0.05 \mathrm{~m}^2\) :
\(
\begin{gathered}
\mathcal{E}=800 \times \frac{\left(5 \times 10^{-5}\right) \times 0.05}{0.1} \\
\mathcal{E}=800 \times \frac{2.5 \times 10^{-6}}{0.1} \\
\mathcal{E}=800 \times 2.5 \times 10^{-5} \\
\mathcal{E}=2 \times 10^{-2} \mathrm{~V}=0.02 \mathrm{~V}
\end{gathered}
\)

Hint

(b) The expression for the EMF is given as:
\(
\mathcal{E}=\int B v d r \dots(1)
\)
Where, \(B=\) magnetic field, \({v}=\) linear velocity, \({dr}=\) elemental radial distance from the centre of the wheel.
Also, \(v=r \omega\) therefore equation (1) becomes,
\(
\begin{aligned}
& \mathcal{E}=\int B \omega r d r=B \omega \int r d r \quad \text { (As } B~ \text { and } \omega \text { are constant) } \\
& \left.\Rightarrow \mathcal{E}=B \omega R^2 / 2 \quad \dots(2) \quad (r~ \text { varies from } 0 \text { to } R\right)
\end{aligned}
\)
Given:
\(
\begin{aligned}
& \mathrm{R}=0.5 \mathrm{~m}, \mathrm{~B}=0.1 \mathrm{~T}, \omega=10 \mathrm{~rad} / \mathrm{s} . \\
& \therefore \mathcal{E}=0.125 \mathrm{~V} . \quad \text { (from equation (2)) }
\end{aligned}
\)

Hint

(d) Magnetic potential energy stored in an inductor is given by
\(
\begin{aligned}
& U=\frac{1}{2} L I^2 \Rightarrow 25 \times 10^{-3}=\frac{1}{2} \times L \times\left(60 \times 10^{-3}\right)^2 \\
& L=\frac{25 \times 2 \times 10^6 \times 10^{-3}}{3600}=\frac{500}{36}=13.89 \mathrm{H}
\end{aligned}
\)

Hint

(b) Given \(n=2 \times 10^4, I=4 \mathrm{~A}\)
Initially \(I=0 \mathrm{~A}\)
\(
\therefore \quad B_i=0 \text { or } \phi_i=0
\)
Finally, the magnetic field at the centre of the solenoid is given as
\(
B_f=\mu_0 n I=4 \pi \times 10^{-7} \times 2 \times 10^4 \times 4=32 \pi \times 10^{-3} \mathrm{~T}
\)
Final magnetic flux through the coil is given as
\(
\begin{aligned}
& \phi_f=N B A=100 \times 32 \pi \times 10^{-3} \times \pi \times(0.01)^2 \\
& \phi_f=32 \pi^2 \times 10^{-5} \mathrm{~T} \mathrm{~m}^2
\end{aligned}
\)
Induced charge, \(q=\frac{|\Delta \phi|}{R}=\frac{\left|\phi_f-\phi_i\right|}{R}=\frac{32 \pi^2 \times 10^{-5}}{10 \pi^2}\)
\(
=32 \times 10^{-6} \mathrm{C}=32 \mu \mathrm{C}
\)

Hint

(d) Emf generated in loop 1,
\(
\begin{aligned}
& \varepsilon_1=-\frac{d \phi}{d t}=-\frac{d}{d t}(\vec{B} \cdot \vec{A})=-\frac{d}{d t}(B A)=-A \times \frac{d B}{d t} \\
& \varepsilon_1=-\left(\pi r^2 \frac{d B}{d t}\right)
\end{aligned}
\)
\(\left(\because A=\pi r^2\right.\) because \(\frac{d B}{d t}\) is restricted upto radius \(r\).)
Emf generated in loop 2,
\(
\varepsilon_2=-\frac{d}{d t}(B A)=-\frac{d}{d t}(0 \times A)=0
\)

Hint

(c) Here, \(N=1000, I=4 \mathrm{~A}, \phi_0=4 \times 10^{-3} \mathrm{~Wb}\)
Total flux linked with the solenoid, \(\phi=N \phi_0\)
\(
=1000 \times 4 \times 10^{-3} \mathrm{~Wb}=4 \mathrm{~Wb}
\)
Since, \(\phi=L I\)
\(\therefore\) Self-inductance of solenoid, \(L=\frac{\phi}{I}=\frac{4 \mathrm{~Wb}}{4 \mathrm{~A}}=1 \mathrm{H}\)

Hint

(b)

When the electron moves from \(X\) to \(Y\), the flux linked with the coil \(a b c d\) (which is into the page) will first increase and then decrease as the electron passes by. So the induced current in the coil will be first anticlockwise and will reverse its direction (i.e., will become clockwise) as the electron goes past the coil.

Note: Presence of magnetic flux is not enough. The amount of magnetic flux linked with the coil must change in order to produce any induced emf in the coil.

Hint

(d) Emf induced in side 1 of frame \(\mathrm{e}_1=\mathrm{B}_1 \mathrm{~V} \ell\)
\(
\mathrm{B}_1=\frac{\mu_{\mathrm{o}} \mathrm{I}}{2 \pi(\mathrm{x}-\mathrm{a} / 2)}
\)
Emf induced in side 2 of frame \(e_2=B_2 V \ell\)
\(
\mathrm{B}_2=\frac{\mu_0 \mathrm{I}}{2 \pi(\mathrm{x}+\mathrm{a} / 2)}
\)

Emf induced in square frame
\(\mathrm{e}=\mathrm{B}_1 \mathrm{~V} \ell-\mathrm{B}_2 \mathrm{~V} \ell\)
\(
\begin{aligned}
& =\frac{\mu_0 \mathrm{I}}{2 \pi(\mathrm{x}-\mathrm{a} / 2)} \ell \mathrm{V}-\frac{\mu_0 \mathrm{I}}{2 \pi(\mathrm{x}+\mathrm{a} / 2)} \ell \mathrm{V} \\
& \text { or, } \mathrm{e} \propto \frac{1}{(2 \mathrm{x}-\mathrm{a})(2 \mathrm{x}+\mathrm{a})}
\end{aligned}
\)

Hint

(d) Motional emf induced in the semicircular ring \(P Q R\) is equivalent to the motional emf induced in the imaginary conductor \(P R\).
\(
\text { i.e., } \varepsilon_{P Q R}=\varepsilon_{P R}=B v l=B v(2 r) \quad(l=P R=2 r)
\)
Therefore, potential difference developed across the ring is \(2 r B v\) with \(R\) at higher potential.

Hint

(c) As the iron rod is inserted, the magnetic field inside the coil magnetizes the iron increasing the magnetic field inside it. Hence. the inductance of the coil increases. Consequently, the inductive reactance of the coil increases. As a result, a larger fraction of the applied AC voltage appears across the lnductor, leaving less voltage across the bulb. Therefore’ the brightness of the light bulb decreases.

In an alternating current (AC) circuit with a coil of self-inductance \(L\) connected in series with a bulb \(B\) and an AC source, the brightness of the bulb decreases when the inductance of the coil increases. This is because the coil introduces inductive reactance ( \(X_L=2 \pi f L\) ), where \(f\) is the frequency of the AC source.

Note: As the inductance \(L\) increases, the inductive reactance increases, which reduces the current flowing through the circuit. Since the brightness of the bulb is directly proportional to the current, a reduction in current causes a decrease in the brightness of the bulb.

So, the correct condition for the brightness of the bulb to decrease is when the inductance \(L\) increases, leading to an increase in the inductive reactance and a corresponding decrease in the current.

Hint

(a) This is the case of periodic EMI

From graph, it is clear that direction is changing once in \(\frac{1}{2}\) cycle.

Hint

(b) Here, \(I=2.5 \mathrm{~A}, L=5 \mathrm{H}\) Magnetic flux linked with the coil is
\(
\phi_B=L I=(5 \mathrm{H})(2.5 \mathrm{~A})=12.5 \mathrm{~Wb}
\)

Hint

(a)

\(
\text { Here, } \phi=50 t^2+4 \mathrm{~Wb}, R=400 \Omega
\)
Induced emf, \(\varepsilon=-\frac{d \phi}{d t}=-\frac{d}{d t}\left(50 t^2+4\right)=-100 t \mathrm{~V}\)
At \(t=2 \mathrm{~s}, \varepsilon=-200 \mathrm{~V} ;|\varepsilon|=200 \mathrm{~V}\)
Induced current in the coil at \(t=2 \mathrm{~s}\) is
\(
I=\frac{|\varepsilon|}{R}=\frac{200 \mathrm{~V}}{400 \Omega}=\frac{1}{2} \mathrm{~A}=0.5 \mathrm{~A}
\)

Hint

(d) \(|V|=\left|-L \frac{d I}{d t}\right|\) \(|V| \propto\) slope of I-t graph

Hint

(a) The charge through the coil = area of current-time \((i-t)\) graph
\(
\begin{aligned}
& q=\frac{1}{2} \times 0.1 \times 4=0.2 \mathrm{C} \\
& q=\frac{\Delta \phi}{R} \because \text { Change in flux }(\Delta \phi)=q \times R \\
& q=0.2=\frac{\Delta \phi}{10} \\
& \Delta \phi=2 \text { Weber }
\end{aligned}
\)

Hint

\(
\text { (a) Induced emf, } e=-L \frac{d i}{d t}
\)
For \(0 \leq t \leq \frac{T}{4}\),
\(i-t\) graph is a straight line with positive constant slope.
\(
\begin{aligned}
& \therefore \frac{d i}{d t}=\text { constant } \\
& \Rightarrow e=-\mathrm{ve} \text { and constant }
\end{aligned}
\)
For \(\frac{T}{4} \leq t \leq \frac{T}{2}\)
\(i\) is constant \(\therefore \frac{d i}{d t}=0 \Rightarrow e=0\)
For \(\frac{T}{2} \leq t \leq \frac{3 T}{4}\)
\(i-t\) graph is a straight line with negative constant slope.
\(
\therefore \frac{d i}{d t}=\text { constant }
\)
\(\Rightarrow e=+\mathrm{ve}\) and constant
For \(\frac{3 T}{4} \leq t \leq T\)
\(i\) is zero \(\therefore \frac{d i}{d t}=0 \Rightarrow e=0\)

Alternate:

(a) \(\mathrm{e}=-\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}\)
During 0 to \(\frac{\mathrm{T}}{4}, \frac{\mathrm{di}}{\mathrm{dt}}=\) const.
\(
\therefore \mathrm{e}=-\mathrm{ve}
\)
During \(\frac{\mathrm{T}}{4}\) to \(\frac{\mathrm{T}}{2}, \frac{\mathrm{di}}{\mathrm{dt}}=0\)
\(
\therefore \mathrm{e}=0
\)
During \(\frac{\mathrm{T}}{2}\) to \(\frac{3 \mathrm{~T}}{4}, \frac{\mathrm{di}}{\mathrm{dt}}=\) const.
\(
\therefore \mathrm{e}=+\mathrm{ve}
\)
Thus graph given in option (a) represents the variation of induced emf with time.

Hint

(b) Here, Magnetic field, \(B=0.025 \mathrm{~T}\)
Radius of the loop, \(r=2 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m}\) Constant rate at which radius of the loop shrinks,
\(
\frac{d r}{d t}=1 \times 10^{-3} \mathrm{~m} \mathrm{~s}^{-1}
\)
Magnetic flux linked with the loop is
\(
\phi=B A \cos \theta=B\left(\pi r^2\right) \cos 0^{\circ}=B \pi r^2
\)
The magnitude of the induced emf is
\(
\begin{aligned}
& |\varepsilon|=\frac{d \phi}{d t}=\frac{d}{d t}\left(B \pi r^2\right)=B \pi 2 r \frac{d r}{d t} \\
& =0.025 \times \pi \times 2 \times 2 \times 10^{-2} \times 1 \times 10^{-3} \\
& =\pi \times 10^{-6} \mathrm{~V}=\pi \mu \mathrm{V}
\end{aligned}
\)

Hint

(a) Induced emf in the loop is given by \(\mathrm{e}=-\mathrm{B} \cdot \frac{\mathrm{dA}}{\mathrm{dt}}\) where A is the area of the loop.
\(
\mathrm{e}=-\mathrm{B} \cdot \frac{\mathrm{~d}}{\mathrm{dt}}\left(\pi \mathrm{r}^2\right)=-\mathrm{B} \pi 2 \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}}
\)
\(
\mathrm{r}=2 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m}
\)
\(
\mathrm{dr}=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}
\)
\(
\mathrm{dt}=1 \mathrm{~s}
\)
\(
\begin{aligned}
& \mathrm{e}=-0.04 \times 3.14 \times 2 \times 2 \times 10^{-2} \times \frac{2 \times 10^{-3}}{1} \mathrm{~V} \\
& =0.32 \pi \times 10^{-5} \mathrm{~V} \\
& =3.2 \pi \times 10^{-6} \mathrm{~V} \\
& =3.2 \pi \mu \mathrm{~V}
\end{aligned}
\)

Hint

(a) The induced emf will remain constant only in the case of rectangular and square loops. In case of the circular and the elliptical loops, the rate of change of area of the loops during their passage out of the field is not constant, hence induced emf will not remain constant for them.

Hint

(c) Net flux \(N \phi=L i\)
Flux per turn \(=4 \times 10^{-3} \mathrm{~Wb}, i=2 \mathrm{~A}\)
Self Inductance \(L=\frac{N \phi}{i}=\frac{4 \times 10^{-3} \times 500}{2}=1 \text { H }\)

Hint

(a) \(\phi=B A \cos \theta\)

\(
\begin{aligned}
& \therefore \phi=\frac{1}{\pi} \times \pi(0.2)^2 \times \cos 60^{\circ} \\
& =(0.2)^2 \times \frac{1}{2}=0.02 \mathrm{~Wb}
\end{aligned}
\)

Hint

(b) 

\(
\begin{aligned}
&\text { Here, } n_P=50, n_s=1500\\
&\begin{aligned}
& \phi_P=\phi_0+4 t \\
& E_P=\frac{d \phi_P}{d t}=\frac{d}{d t}\left(\phi_0+4 t\right)=4 \\
& \text { As } \frac{E_s}{E_P}=\frac{n_s}{n_P} \\
& \therefore E_s=\frac{n_s}{n_P} \times E_P=\frac{1500}{50} \times 4=120 \mathrm{~V}
\end{aligned}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (c)  Mutual inductance between coils is }\\
&\begin{aligned}
M & =K \sqrt{L_1 L_2}=1 \sqrt{2 \times 10^{-3} \times 8 \times 10^{-3}} \quad(\because K=1) \\
& =4 \times 10^{-3}=4 \mathrm{mH}
\end{aligned}
\end{aligned}
\)

Hint

(a) \(\text { Work done due to a charge } W=Q V\)

Hint

(a) Induced emf is given by \(V=\frac{\Delta \phi}{\Delta t}\) current \((i)=\frac{Q}{\Delta t} \Rightarrow \frac{\Delta \phi}{\Delta t} \times \frac{1}{R}=\frac{Q}{\Delta t}\)
[where \(Q\) is total charge in time \(\Delta t\) ]
\(
\Rightarrow \quad Q=\frac{\Delta \phi}{R}
\)

Hint

\(
\begin{aligned}
&\text { (a) } L=2 \mathrm{mH}, i=t^2 \mathrm{e}^{-t}\\
&\begin{aligned}
& E=-L \frac{d i}{d t}=-L\left[-t^2 e^{-t}+2 t e^{-t}\right] \\
& \text { when } E=0 \\
& -e^{-t} t^2+2 t e^{-t}=0 \\
& \text { or, } 2 t e^{-t}=e^{-t} t^2 \\
& \Rightarrow t=2 \text { sec. }
\end{aligned}
\end{aligned}
\)

Hint

(a) To find the work done by the source to establish a current of 5 A in an inductor coil with inductance \(L=0.04 \mathrm{H}\), we can use the formula for the work done in an inductor:
\(
W=\frac{1}{2} L I^2
\)
1. Identify the given values:
Inductance \(L=0.04 \mathrm{H}\)
Current \(I=5 \mathrm{~A}\)
2. Substitute the values into the work done formula:
\(
\begin{aligned}
& W=\frac{1}{2} \times L \times I^2 \\
& W=\frac{1}{2} \times 0.04 \mathrm{H} \times(5 \mathrm{~A})^2
\end{aligned}
\)
3. Calculate \(I^2\) :
\(
I^2=5^2=25
\)
4. Substitute \(I^2\) back into the equation:
\(
W=\frac{1}{2} \times 0.04 \times 25
\)
5. Calculate \(W\) :
\(
W=\frac{1}{2} \times 1=0.5 \mathrm{~J}
\)

Hint

(c) Faraday’s law states that the induced EMF \(\mathcal{E}\) in a loop is proportional to the rate of change of the magnetic flux \(\Phi_B\) through the loop:
\(
\mathcal{E}=-\frac{d \Phi_B}{d t}
\)
where \(\Phi_B=B A \cos (\theta)\) is the magnetic flux, with:
\(B\) being the magnetic field,
\(A\) being the area of the coil,
\(\theta\) being the angle between the magnetic field \(B\) and the normal to the coil.
Initially, the plane of the coil is parallel to the magnetic field, so \(\theta=0\) and \(\cos (0)=1\), meaning the magnetic flux is at its maximum:
\(
\Phi_B=B A
\)
At time \(\Delta t\), the plane of the coil is rotated to become perpendicular to the magnetic field, so \(\theta=\) \(90^{\circ}\), and \(\cos \left(90^{\circ}\right)=0\). This means that the final flux is zero:
\(
\Phi_B^{\prime}=0
\)
The change in magnetic flux \(\Delta \Phi_B\) is the difference between the initial and final flux:
\(
\Delta \Phi_B=B A-0=B A
\)
Induced EMF:
The induced EMF is given by the rate of change of magnetic flux. The flux change occurs over a time interval \(\Delta t\), so:
\(
\mathcal{E}=\frac{\Delta \Phi_B}{\Delta t}=\frac{B A}{\Delta t}
\)
Current and Charge Flow:
By Ohm’s law, the current \(I\) induced in the coil is:
\(
I=\frac{\mathcal{E}}{R}=\frac{B A}{R \Delta t}
\)
The total charge \(Q\) that flows during this time \(\Delta t\) is given by:
\(
Q=I \Delta t=\frac{B A}{R \Delta t} \times \Delta t=\frac{B A}{R}
\)
Notice that the charge \(Q\) is independent of the time interval \(\Delta t\). Therefore, the charge that flows in the time \(\Delta t\) is proportional to \(\Delta t^0\).
Conclusion:
The charge that flows in time \(\Delta t\) depends on time as \(Q \propto(\Delta t)^0\).

Hint

(b) \(\text { As, }|\varepsilon|=M \frac{d I}{d t}\)
\(
=M \frac{d}{d t}\left(I_0 \sin \omega t\right)=M I_0 \omega \cos \omega t
\)
\(
\therefore \quad \varepsilon_{\max }=0.005 \times 10 \times 100 \pi \times 1=5 \pi
\)

Hint

(c) When the magnet is dropped through the ring, an induced current is developed into the ring in the direction opposing the motion of magnet (Lenz’s law). Therefore this induced current decreases the acceleration of bar magnet.

Hint

(d) Length of conductor \((l)=0.4 \mathrm{~m}\), Speed \((v)=7 \mathrm{~m} / \mathrm{s}\) and magnetic field \((B)=0.9 \mathrm{~Wb} / \mathrm{m}^2\) Induced e.m.f. \((\varepsilon)=B l v=0.9 \times 0.4 \times 7=2.52 \mathrm{~V}\).

Hint

(c) Initial current \(\left(\mathrm{I}_1\right)=10 \mathrm{~A}\); Final current \(\left(\mathrm{I}_2\right)\) \(=0 ;[latex] Time [latex](\mathrm{t})=0.5 \mathrm{sec}\) and induced e.m.f. \((\varepsilon)=\) 220 V.
\(
\begin{aligned}
&\begin{aligned}
& \text { Induced e.m.f. }(\varepsilon) \\
& =-L \frac{d I}{d t}=-L \frac{\left(I_2-I_1\right)}{t}=-L \frac{(0-10)}{0.5}=20 \mathrm{~L} \\
& \text { or, } \mathrm{L}=\frac{220}{20}=11 \mathrm{H}
\end{aligned}\\
&\text { [where L = Self inductance of coil] }
\end{aligned}
\)

Hint

(c) \(L=\frac{N \phi}{i} ; \phi=B A ; B=\mu_0 n i=\frac{\mu_0 N i}{l}\)
\(
L=\frac{\mu_0 N^2}{l} A=\mu_0 n^2 A l
\)
where \(n\) is the number of turns per unit length \(L \propto N^2\)

Hint

(b)

\(
\begin{aligned}
& \varepsilon=-L \frac{d i}{d t} \\
& L=\frac{-\varepsilon}{\frac{d i}{d t}}=\frac{-5 \times 10^{-3}}{(2-3)} \mathrm{H}=5 \mathrm{mH}
\end{aligned}
\)

Hint

(c) \({q}=\int i d t=\frac{1}{R} \int \varepsilon d t=\left(\frac{-d \phi}{d t}\right) \frac{1}{R} \int d t=\frac{1}{R} \int d \phi\)

Hence total charge induced in the conducting loop depend upon the total change in magnetic flux. As the emf or \(i R\) depends on rate of change of \(\phi\), charge induced depends on change of flux.

Hint

(c)

\(
\begin{aligned}
: i & =\frac{\varepsilon}{R}=\frac{\frac{N A d B}{d t}}{R} \\
& =\frac{20 \times\left(25 \times 10^{-4}\right) \times 1000}{100}=0.5 \mathrm{~A}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { (b) } \varepsilon=\frac{-\left(\phi_2-\phi_1\right)}{t}=\frac{-(0-N B A)}{t}=\frac{N B A}{t} \\
& t=\frac{N B A}{\varepsilon}=\frac{50 \times 2 \times 10^{-2} \times 10^{-2}}{0.1}=0.1 \mathrm{~s}
\end{aligned}
\)

Hint

\(
\text { (c) } E=\frac{1}{2} L i^2=\frac{1}{2} \times\left(100 \times 10^{-3}\right) \times 1^2=0.05 \mathrm{~J}
\)

Hint

(d) Self inductance of a solenoid \(=\frac{\mu n^2 A}{\ell}\)

So, self induction \(\propto n^2\)
So, inductance becomes 4 times when n is doubled.

Hint

(a) According to Faraday’s law, it is the conservation of energy.

Hint

(a) \(|\varepsilon|=L \frac{d i}{d t}\)
Given that, \(L=40 \times 10^{-3} \mathrm{H}, d i=11 \mathrm{~A}-1 \mathrm{~A}=10 \mathrm{~A}\) and \(d t=4 \times 10^{-3} \mathrm{~s}\)
\(
\therefore|\varepsilon|=40 \times 10^{-3} \times\left(\frac{10}{4 \times 10^{-3}}\right)=100 \mathrm{~V}
\)

Hint

(c) An inductor stores energy in its magnetic field.

Hint

\(
\begin{aligned}
&\text { (c) } e=\frac{d \phi}{d t} ; i=\frac{e}{R}=\frac{1}{R} \frac{d \phi}{d t}\\
&\begin{aligned}
\text { Total charge induced } & =\int i d t=\int \frac{1}{R} \frac{d \phi}{d t} d t \\
& =\frac{1}{R} \int_{\phi_1}^{\phi_2} d \phi=\frac{1}{R}\left(\phi_2-\phi_1\right)
\end{aligned}
\end{aligned}
\)

Hint

(c) When a circuit is broken, the induced e.m.f. is largest. 

Explanation: When contact is broken, current in the circuit decreases. To oppose this inductor releases current and emf is induced. This induced emf is having high value hence bulb will become suddenly bright.

Hint

(a)  Eddy currents are produced when a metal is kept in a varying magnetic field.

Hint

(c) \(I_0=\frac{E_0}{R}=N \frac{B A \omega}{R}\)
Given, \(N=1, B=10^{-2} \mathrm{~T}, A=\pi(0.3)^2 \mathrm{~m}^2, R=\pi^2 \Omega\)
\(
f=(200 / 60) \text { and } \omega=2 \pi(200 / 60)
\)
Substituting these values and solving, we get
\(
I_0=6 \times 10^{-3} \mathrm{~A}=6 \mathrm{~mA}
\)

Hint

(c) The magnetic moment of an electron in its lowest energy state, according to the model, is \(\frac{h e}{2 \pi m}\). This corresponds to option (c) from the given choices.
Explanation:
Quantization Condition: The flux passing through the electron’s orbit is quantized, meaning it’s a multiple of \(\frac{h}{e}\). For the lowest energy state, \(n=1\).
Magnetic Flux: Magnetic flux \(\phi\) is given by the product of the magnetic field \(B\) and the area of the orbit, \(\phi=\boldsymbol{B} \cdot \boldsymbol{A}\).
Quantized Flux: \(\phi=n \frac{h}{e}\), so for the lowest state, \(A=\frac{h}{e B}\).
Area of Orbit: The area of a circular orbit is \(A=\pi r^2\), so \(\pi r^2=\frac{h}{e B}\).
Radius of Orbit: The radius of the electron’s orbit can be found from the area,
\(
r=\sqrt{\frac{h}{e B \pi}}
\)
Velocity and Magnetic Force: The centripetal force is provided by the magnetic force, \(\frac{m v^2}{r}=e v \boldsymbol{B}\). Hence, \(v=\frac{e \boldsymbol{B} r}{m}\).
Magnetic Moment: The magnetic moment \(\mu\) of a current loop is \(\mu=I A\) where \(I\) is the current and \(A\) is the area. The current is \(I=\frac{e v}{2 \pi r}\).
Substituting: Substituting for \(v\) and \(A\) in the magnetic moment equation we get: \(\mu=\frac{e v}{2 \pi r} \cdot \pi r^2=\frac{e v r}{2}\). Further substituting \(r=\sqrt{\frac{h}{e B \pi}}\) and \(v=\frac{e B r}{m}\) yields \(\mu=\frac{e}{2} \sqrt{\frac{h}{e B \pi}} \frac{e B}{m} \sqrt{\frac{h}{e B \pi}}=\frac{h e}{2 \pi m}\).