NEET Practice Q & A

Hint

(d) We have, \(f=\frac{1}{\tau_C}=\frac{1}{R C}, f=\frac{1}{\tau_L}=\frac{R}{L}\)
and \(\omega=\frac{1}{\sqrt{L C}}\) \(\therefore \frac{C}{L}\) does not have dimensions of frequency.

Hint

(b) Induced emf, \(e=-\frac{d \phi}{d t}\)
But, \(e=i R \quad\) and \(\quad i=\frac{d q}{d t}\)
\(
\therefore \quad \frac{d q}{d t} R=-\frac{d \phi}{d t} \text { or } d q=-\frac{d \phi}{R}
\)
So, induced charge in a coil is independent of time taken to change the flux.

Hint

(b) The value of self induction of coil increases when soft iron core is inserted in the coil, hence current decreases due to increase of its reactance. So, intensity of bulb decreases.

Hint

(b) The magnitude of induced emf is directly proportional to the rate of change of magnetic flux. Also, induced charge is \(d q=-\frac{d \phi}{R}\), which is independent of time.
\(\therefore\) Induced emf and induced charge will be more in first case and equal in both case, respectively.

Hint

\(
\begin{aligned}
&\text { (a) Induced current is given by }\\
&\begin{aligned}
i & =\frac{e}{R}=-\frac{N(d \phi / d t)}{R}=-\frac{N}{R}\left(\frac{d \phi}{d t}\right)=\frac{-N}{R} \cdot \frac{d B}{d t} \cdot A \\
& =\frac{10 \times 10^8 \times 10^{-4} \times 10^{-4} \times 10}{20}=5 \mathrm{~A}
\end{aligned}
\end{aligned}
\)

Hint

\(
\text { (c) Applying Lenz’s law, the loop } B \text { is repelled by loop } A \text {. }
\)

Hint

\(
\begin{aligned}
&\text { (d) Induced emf, }\\
&\begin{aligned}
e & =-N \frac{d \phi}{d t} \\
|e| & =N \frac{d \phi}{d t}=N \frac{d}{d t}(B A \cos \theta)=N \frac{d B}{d t} \cdot A \cos \theta \\
& =500 \times 1 \times\left(10 \times 10^{-2}\right)^2 \cos 0^{\circ}=5 \mathrm{~V}
\end{aligned}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (b) Induced emf, } e=-\frac{d \phi}{d t}\\
&|e|=\frac{d}{d t}(B A)=A \frac{d B}{d t}=2 \times \frac{(4-1)}{2}=3 \mathrm{~V}
\end{aligned}
\)

Hint

(a) The two loops will attract each other.
Explanation: When the current in the first loop changes due to temperature variations, it induces a current in the second loop. According to Lenz’s Law, the induced current will oppose the change in magnetic flux. Since the current in the first loop is decreasing, the induced current in the second loop will be in the same direction as the original current in the first loop. This means the loops will experience an attractive force due to the same direction currents.

Key points:
Lenz’s Law:
The induced current always opposes the change in the magnetic flux that produced it. ©

Attraction:
When currents in two loops are in the same direction, they attract each other.
Repulsion:

Hint

(c) \(\phi=\mathbf{B} \cdot \mathbf{A}=B A \cos 90^{\circ}=0\). Therefore, \(d \phi=0\) and \(i=\frac{d q}{d t}=\frac{1}{R} \frac{d \phi}{d t}=0\)

Hint

(a) The magnet gets repelled. Therefore, north pole is formed on the top side and hence current will flow in anti-clockwise direction.

Hint

(c) Current in the circuit, \(i=V / R=10 / 2=5 \mathrm{~A}\)
\(\therefore \quad\) Magnetic energy stored in the coil,
\(
U=\frac{1}{2} L i^2=\frac{1}{2} \times 2 \times(5)^2=25 \mathrm{~J}
\)

Hint

(b) Time constant, \(\tau_L=\frac{L}{R}=\frac{5}{10}=\frac{1}{2}\) We have, \(\quad i=i_0\left(1-e^{-t / \tau_L}\right)\)
\(
\therefore \quad \frac{i_{\infty}}{i_1}=\frac{i_0}{i_0\left(1-e^{-t / \tau_L}\right)}
\)
Substituting, \(t=1 \mathrm{~s}\) and \(\tau_L=\frac{1}{2} \mathrm{~s}\), we get
\(
\frac{i_{\infty}}{i_1}=\frac{e^2}{e^2-1}
\)

Hint

\(
\begin{aligned}
&\text { (a) Charge induced in coil is given as }\\
&\begin{aligned}
d q & =\frac{d \phi_B}{R}=i d t=\text { Area under } i-t \text { graph } \\
d \phi_B & =(\text { Area under } i-t \text { graph }) R \\
& =\frac{1}{2} \times 4 \times 0.1 \times 10=2 \mathrm{~Wb}
\end{aligned}
\end{aligned}
\)

Hint

(d) Here, inductor \(L\) and bulb \(B_2\) make a \(L R\) circuit, whereas resistor \(R\) and bulb \(B_1\) make a simple \(R\) circuit.
So, \(B_1\) dies out promptly but \(B_2\) with some delay.

Hint

\(
\text { (d) We have, } \Delta q=-\frac{N}{R} \Delta \phi \text { and } \Delta \phi=\Delta B \cdot A \cos \theta
\)
\(
\begin{aligned}
&\therefore \quad 32 \times 10^{-6}=-\frac{100}{(160+40)}(0-B) \times \pi \times\left(6 \times 10^{-3}\right)^2 \times \cos 0^{\circ}\\
&\therefore \text { Intensity of magnetic field, } B=0.566 \mathrm{~T}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (d) }\\
&\begin{aligned}
& \text { Induced emf, } e=\left|\frac{-N B A\left(\cos \theta_2-\cos \theta_1\right)}{\Delta t}\right| \\
& \quad=\left|-\frac{800 \times 4 \times 10^{-5} \times 0.05\left(\cos 90^{\circ}-\cos 0^{\circ}\right)}{0.1}\right|=0.016 \mathrm{~V}
\end{aligned}
\end{aligned}
\)

Hint

(a) Magnetic flux, \(\phi=B \cdot A\)
Change in flux, \(d \phi=B \cdot d A\)
\(
=0.05(101-100) \times 10^{-4}=5 \times 10^{-6} \mathrm{~Wb}
\)
Charge, \(d q=\frac{d \phi}{R}=\frac{5 \times 10^{-6}}{2}=2.5 \times 10^{-6} \mathrm{C}\)

Hint

\(
\begin{aligned}
&\text { (b) For growth of current in } L-R \text { circuit, current is given by }\\
&\begin{aligned}
i & =i_0\left(1-e^{-R t / L}\right) \\
\therefore \quad \frac{d i}{d t} & =-i_0\left(-\frac{R}{L}\right) e^{-R t / L}=\frac{i_0 R}{L} e^{-R t / L}
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
&\text { At the instant of closing the switch, } \frac{d i}{d t}=\frac{i_0 R}{L}=\frac{E}{L}\\
&\begin{array}{ll}
\Rightarrow & 4=\frac{E}{20} \\
\therefore & E=20 \times 4=80 \mathrm{~V}
\end{array}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (d) Peak value of emf generated in generator is given by }\\
&\begin{aligned}
e_0 & =\omega N B A=(2 \pi v) N B A \\
& =2 \times 3.14 \times 100 \times 5000 \times 0.2 \times 0.25=157 \mathrm{kV}
\end{aligned}
\end{aligned}
\)

Hint

(d) If electron is moving from left to right, the flux linked with the loop (which is into the page) will first increase and then decrease as the electron passes through it. So, the induced current in the loop will be first anti-clockwise and will change direction as the electron passes through.

Hint

(d)  \(\mathbf{B}, \mathbf{v}[latex] and [latex]\mathbf{I}\) should be mutually perpendicular.
So, the conductor should more in the direction perpendicular to the plane of \(l\) and \(B\), i.e. in the direction of \(M\).

Hint

(c) Induced emf, \(e=B v l\)
\(
\begin{array}{ll}
\therefore & e \propto v \\
\text { Also, } & v=0+g t \\
\therefore & e \propto g t
\end{array}
\)
So, potential difference between its two ends will increase with time.

Hint

(a) In the given circuit, three inductances are in parallel, their equivalent inductance is given by
\(
\frac{1}{L_{\mathrm{eq}}}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3} \text { or } L_{\mathrm{eq}}=\frac{3}{3}=1 \mathrm{H}
\)

Hint

\(
\begin{aligned}
&\text { (d) }\\
&\begin{gathered}
\therefore \text { In parallel, } L_P=\frac{L_1 L_2}{L_1+L_2} \\
\left(L_1+L_2\right) 1.5=L_1 L_2 \dots(i)
\end{gathered}
\end{aligned}
\)
\(
\begin{array}{lc}
\text { and in series, } & L_S=L_1+L_2=8 \\
\therefore & L_1 L_2=12 \text { [from Eq. (i)] }
\end{array}
\)
\(
\begin{aligned}
\left(L_1-L_2\right)^2= & \left(L_1+L_2\right)^2-4 L_1 L_2 \\
L_D & =\sqrt{(8)^2-4 \times 12} \\
L_D & =4 \mathrm{H}
\end{aligned}
\)

Hint

(c) Time constant, \(\tau=\frac{L}{R}\)
It is the time interval during which the current after opening an inductive circuit falls to \(37 \%\) of its maximum value.

Explanation:

The current decay equation is \(I=I_0 e^{-\frac{R t}{L}}\). Substitute the value of \(I: 0.37 I_0=I_0 e^{-\frac{R t}{L}}\).
Recognize that 0.37 is approximately \(e^{-1}\). So, \(e^{-1}=e^{-\frac{R t}{L}}\).
Equate the exponents: \(-1=-\frac{R t}{L}\).
Solve for \(t: t=\frac{L}{R}\).

Hint

(c) Time constant, \(\tau=\frac{L}{R}=2.0 \times 10^{-3} \mathrm{~s}\) and \(\frac{L}{R+90}=0.5 \times 10^{-3} \mathrm{~s}\)
Solving these two equations, we get
\(
L=60 \mathrm{mH} \text { and } R=30 \Omega
\)

Hint

(c) Four resistances form a balanced Wheatstone bridge. Therefore, energy stored in the coil is zero (current is 0 through the coil).

Hint

(d) The whole current will pass through the inductor, since \(R=0\). So, the current in \(10 \Omega\) resistor is zero.

Explanation: In the steady state (after a long time), the inductor acts as a short circuit. This means the inductor offers zero resistance to the current.  Since the inductor is in parallel with the \(10 \Omega\) resistor and offers zero resistance, all the current will flow through the inductor. No current will flow through the \(10 \Omega\) resistor.

Hint

\(
\text { (b) Power, } P=F v=i l B v=\left(\frac{B v l}{R}\right) l B v=\frac{B^2 l^2 v^2}{R}
\)

Hint

\(
\begin{aligned}
&\text { (a) } \therefore \text { Magnetic flux, }\\
&\begin{aligned}
\phi & =B A \cos \theta=0.2 \sin (300 t) \times 5 \times 10^{-4} \times \cos 60^{\circ} \\
& =5 \times 10^{-5} \sin (300 t)
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
\therefore \text { emf, } e & =\frac{d \phi}{d t}=5 \times 10^{-5} \times 300 \cos (300 t) \\
& =1.5 \times 10^{-2} \cos \left(300 \times \frac{\pi}{900}\right) \quad\left(\because t=\frac{\pi}{900} \mathrm{~s}\right) \\
& =1.5 \times 10^{-2} \cos (\pi / 3)=0.75 \times 10^{-2}=7.5 \times 10^{-3} \mathrm{~V}
\end{aligned}
\)

Hint

(c) Since, the magnetic field is uniform, therefore there will be no change in flux, hence no current will be induced.

Hint

(d) An EMF will not be induced in the coil if the magnetic flux through the coil remains constant. This occurs when the coil is stationary in a uniform magnetic field and the plane of the coil is perpendicular to the field, or when the coil rotates at a constant angular velocity such that the rate of change of flux through the coil is zero.

Hint

(b) Charge, \(\Delta Q=\frac{\Delta \phi}{R}=\frac{N \times B A}{R}\)
\(\therefore\) Magnetic flux density,
\(
B=\frac{\Delta Q \cdot R}{N A}=\frac{2 \times 10^{-4} \times 80}{40 \times 4 \times 10^{-4}}=1 \mathrm{~Wb} / \mathrm{m}^2
\)

Hint

(b) We have, \(L \propto N^2\)
\(
\frac{L_B}{L_A}=\left(\frac{N_B}{N_A}\right)^2
\)
Self-inductance, \(L_B=\left(\frac{500}{600}\right)^2 \times 108=75 \mathrm{mH}\)

Hint

(d) By Fleming’s right hand rule, field due to wire inwards and perpendicular to the plane. This field keeps on decreasing as distance from wire increases. So, potential at A will be higher than that at B.

Alternate: The potential at end A of the rod will be higher than at end B . This is because the magnetic field generated by the current in the wire is stronger closer to the wire. As the rod moves parallel to the wire, the side of the rod closer to the wire experiences a stronger magnetic force on its free electrons, resulting in a higher potential at that end.

Hint

(a) Mutual inductance between two coils depend on their degree of flux linkage, i.e. the fraction of flux linked with one coil when some current passes through the other coil. In Fig. (A), two coils with their planes are parallel. In this situation, maximum flux passes. So, mutual inductance will be maximum in this case.

Hint

\(
\begin{aligned}
&\text { (d) During growth of current in the coil, }\\
&\begin{aligned}
& & i & =i_0\left(1-e^{-R t / L}\right) \\
\text { For } & & i & =\frac{i_0}{2} \\
\Rightarrow & & \frac{i_0}{2} & =i_0\left(1-e^{-R t / L}\right) \\
\therefore & & t & =0.693 \frac{L}{R} \\
& & & =0.693 \times \frac{300 \times 10^{-3}}{2}=0.1 \mathrm{~s}
\end{aligned}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (c) Growth of current in the circuit is } i=i_0\left(1-e^{-R t / L}\right)\\
&\begin{array}{ll}
\Rightarrow & \frac{d i}{d t}=\frac{d}{d t} i_0-\frac{d}{d t}\left(i_0 e^{-R t / L}\right) \\
\therefore & \frac{d i}{d t}=0-i_0\left(-\frac{R}{L}\right) e^{-R t / L}=\frac{i_0 R}{L} e^{-R t / L}
\end{array}
\end{aligned}
\)
\(
\begin{aligned}
& \text { Initially, } t=0 \\
& \therefore \quad \frac{d i}{d t}=\frac{i_0 R}{L}=\frac{E}{L}=\frac{5}{2}=2.5 \mathrm{As}^{-1}
\end{aligned}
\)

Hint

(a) Just after pressing the key \(K\), the inductor offers infinite resistance. So, net resistance will be \(14 \Omega\).
\(
\therefore \quad I=V / R=\frac{10}{14}=(5 / 7) \mathrm{A}
\)

Hint

(a) At \(t=0\), current will flow from the capacitor (capacitor behaves as short at \(t=0\)) and at \(t=\infty\) current will flow from the inductor(inductor behaves as short in the steady state). So, current in both cases will be 1.5 A.

Hint

(c) Applying Kirchhoff’s second law to above circuit,

\(
\begin{gathered}
V_A-i R+15-L \frac{d i}{d t}=V_B \\
V_A-5 \times 1+15-\left(5 \times 10^{-3}\right)\left(-10^3\right)=V_B \\
V_B-V_A=15 \mathrm{~V}
\end{gathered}
\)

Hint

(c) At \(t=0\), induced emf \(e\) is maximum and is equal to applied emf, but current \(i\) is zero.
As, \(i=\frac{E}{R}\left(1-e^{-t R / L}\right)\) and \(V=\frac{L d i}{d t}=L \cdot \frac{E}{R}\left(\frac{R}{L}\right) e^{-t R / L}=E e^{-t R / L}\)
So, after lapse of time, current through the circuit increases but induced emf decreases.

Hint

(b) Two coils are said to be magnetically coupled, if full or a part of the flux produced by one link with the other. Let \(L_1\) and \(L_2\) be the self-inductances of the coils and \(M\) be their mutual inductances, then coefficient of coupling \(k=\frac{M}{\sqrt{L_1 L_2}}\).
When \(k=1\), then \(100 \%\) flux produced by one coil links with the other. In this case, mutual inductance between the two is maximum and is given by \(M=\sqrt{L_1 L_2}\)

Hint

(b) Induced emf, \(|e|=\frac{d \phi}{d t}=\frac{B d S}{d t}\)
Now, as the square loop and rectangular loop move out of magnetic field, \(\frac{B d S}{d t}\) is constant, therefore \(|e|\) is constant. But in case of circular and elliptical loops, \(\frac{d S}{d t}\) changes. Therefore, emf does not remain constant.

Hint

(d) When the magnet oscillates in and out of the spring, it induces an EMF, the direction in which EMF is getting induced will be different. Due to the induced EMF, a current will be set up in the coil which will deflect the pointer in the galvanometer in the opposite directions, as the magnet oscillates in and out of the spring an eddy current is set up in it decreases the amplitude of oscillation.
In short, the EMF will be induced in opposite directions, ie: Left and Right, as the magnet oscillates in and out of the spring also the eddy current will reduce the amplitude of oscillation as the time goes on (Damping)

Hint

(d)

Eddy current induced in aluminium plate create damping effect in coil.

Hint

(d) Mutual inductance between two coils in the same plane with their centres coinciding is given by
\(
M=\frac{\mu_0}{4 \pi}\left(\frac{2 \pi^2 R_2^2 N_1 N_2}{R_1}\right) \text { henry } \Rightarrow M \propto \frac{R_2^2}{R_1}
\)

Hint

(b) Power, \(P=\frac{e^2}{R}\), and \(e=-\left(\frac{d \phi}{d t}\right)\), where \(\phi=\) NBS
\(
e=-N S\left(\frac{d B}{d t}\right) . \text { Also, } R \propto \frac{l}{r^2}
\)
where, \(R=\) resistance, \(r=\) radius, \(l=\) length
\(
\begin{array}{ll}
\therefore & P \propto \frac{N^2 r^2}{l} \\
& \frac{P_1}{P_2}=\left(\frac{N_1}{N_2}\right)^2\left(\frac{r_1}{r_2}\right)^2 \frac{l_2}{l_1} \dots(i)
\end{array}
\)
\(
\begin{aligned}
\text { Since, } V_1 & =V_2 \\
\pi r_1^2 l_1 & =\pi r_2^2 l_2 \\
\Rightarrow \quad \frac{l_1}{l_2} & =\left(\frac{r_2}{r_1}\right)^2 \Rightarrow \frac{l_2}{l_1}=\left(\frac{r_1}{r_2}\right)^2 \dots(ii)
\end{aligned}
\)
\(
\begin{aligned}
&\therefore \text { From Eqs. (i) and (ii), we get }\\
&\begin{aligned}
\frac{P_1}{P_2} & =\left(\frac{N_1}{N_2}\right)^2 \cdot\left(\frac{r_1}{r_2}\right)^2 \cdot\left(\frac{r_1}{r_2}\right)^2 \\
& =\left(\frac{N_1}{N_2}\right)^2 \cdot\left(\frac{r_1}{r_2}\right)^4=\left(\frac{N_1}{4 N_1}\right)^2\left(\frac{r_1}{r_1 / 2}\right)^4 \\
& =\frac{1}{4^2} \times 2^4=\frac{1}{16} \times 16=1 \\
\Rightarrow \quad \frac{P_1}{P_2} & =1 \Rightarrow P_1=P_2
\end{aligned}
\end{aligned}
\)

Hint

(b) As the magnet moves towards the coil, the magnetic flux increases (non-linearly). Also, there is a change in polarity of induced emf when the magnet passes on to the other side of the coil. So, correct variation is shown in (b).

Hint

(b) \(i=i_0\left(1-e^{-\frac{R}{L} t}\right)\)
At \(t=0, i=0\).
At \(t \gg \tau(=L / R), i=i_0\).
Thus, \(i\) increases exponentially upto \(i_0\) and then become constant as shown in (b).

Hint

(c) Since, you are decreasing current in the circuit by increasing resistance of the circuit, then induced emf across inductor will support the battery. Hence, net emf of the circuit is greater than 20 V or current in the circuit is more than 4 A .

Hint

(c) After short-circuiting the battery.

Resistance, \(R_{a b}=\frac{R_2\left(R_1+r\right)}{R_2+R_1+r}\)
\(\therefore\) Time constant, \(\tau_L=\frac{L}{R_{\text {net }}}=\frac{L\left(R_1+R_2+r\right)}{\left(R_1+r\right) R_2}\)

Hint

(d) Magnetic lines on circular loop, due to current in wire \(A B\) are tangential to its plane.
\(\therefore\) Magnetic flux, \(\phi_B=0\) or induced emf, \(\frac{d \phi_B}{d t}=0\)

Hint

(a) Induced emf, \(|e|=\frac{\Delta \phi}{\Delta t}=\frac{(2-1)(0.01)}{10^{-3}}=10 \mathrm{~V}\)
\(\therefore\) Current, \(i=\frac{e}{R}=\frac{10}{0.01}=10^3 \mathrm{~A}\)
Heat produced, \(H=\) eit \(=(10)\left(10^3\right)\left(10^{-3}\right)=10 \mathrm{~J}\)

Hint

\(
\begin{aligned}
&\text { (d) As, potential difference, }\\
&V=B v l=\left(2.0 \times 10^{-4}\right)\left(360 \times \frac{5}{18}\right)(50)=1.0 \mathrm{~V}
\end{aligned}
\)

Hint

(a) If the loop remains inside the region of magnetic field, no current will be induced. But, if it moves out of the plane, then the induced current is given by \(i=\frac{B l v}{R}\) and its direction will be clockwise (as per Lenz’s law).

Hint

(b) As, potential difference, 

\(
\begin{aligned}
 e & =B v l=\left(0.2 \times 10^{-4}\right)(20)(1) \mathrm{V} \\
& =0.4 \mathrm{mV}
\end{aligned}
\)

Hint

(c) \(M_{12}=M_{21}\), when rate of change of current is doubled, induced emf also becomes two times.
Thus, emf produced across the first coil is 6 mV.

Hint

(b) Two emfs are induced in the closed circuit each of value \(B l v\). These two emfs are additive, so \(E_{\text {net }}=2 B l v\).

Hint

(d) Magnetic flux,
\(
\begin{aligned}
\phi & =N A B=(25)(\pi)\left(1.8 \times 10^{-2}\right)^2\left[0.2 t-0.05 t^2\right] \\
& =0.025\left(0.2 t-0.05 t^2\right)
\end{aligned}
\)
Induced emf, \(e=\left|\frac{d \phi}{d t}\right|=|0.025(0.2-2 \times 0.05 t)|\)
At
\(
\begin{aligned}
& t=3 \mathrm{~s}, e=0.0025 \mathrm{~V} \\
& i=\frac{e}{R}=0.0016 \mathrm{~A}
\end{aligned}
\)
\(\therefore\) Power dissipation, \(P=e i=4 \times 10^{-6} \mathrm{~W}=4 \mu \mathrm{~W}\)

Hint

(a) In the case (a), the induced current in the ring will be in such a direction that it repels the magnet.
i.e. \(a<g\) and moving towards \(R\).

In case (c), the induced current is such as to attract the magnet.
i.e. \(a<g\) and moving away from \(R\).

Hint

(b) Here, \(e=B v l=(0.1)(1)(0.1)=\frac{1}{100} \mathrm{~V}\)
Net resistance \(=1+\frac{2 \times 3}{2+3}=\frac{11}{5} \Omega\)
\(\therefore\) Current, \(i=\left(\frac{1 / 100}{11 / 5}\right)=\frac{1}{220} \mathrm{~A}\)

Hint

(a) From the figure, \(V_a-i R-L \frac{d i}{d t}=V_b\)
\(
\therefore \quad V_a-V_b=i R+L \frac{d i}{d t}
\)
According to given conditions,
\(
\begin{aligned}
& 8=2 R+L \dots(i) \\
& 4=2 R-L \dots(ii)
\end{aligned}
\)
Solving these two equations, we get
\(\)
R=3 \Omega \text { and } L=2 \mathrm{H}
\(\)

Hint

(b) \(V=L \frac{d i}{d t}\) or \(V \propto L\) \(\left(\because \frac{d i}{d t}=\right.\) constant \()\)
\(
\therefore \frac{V_1}{V_2}=\frac{L_1}{L_2}=4 \quad(\because P=V i=\text { constant })
\)
\(
\begin{array}{ll}
\therefore & i \propto \frac{1}{V} \text { or } \frac{i_1}{i_2}=\frac{V_2}{V_1}=\frac{1}{4} \\
\therefore & \frac{W_1}{W_2}=\frac{L_1}{L_2}\left(\frac{i_1}{i_2}\right)^2=(4)\left(\frac{1}{16}\right)=\frac{1}{4} \left(\because W=\frac{1}{2} L i^2\right) \\
\because & \frac{i_1}{i_2} \neq 4
\end{array}
\)

Hint

(c) Induced emf, \(E=-L \frac{d i}{d t}\) or induced emf \(\propto-\) slope of \(i-t\) graph. So, graph (c) is correct.

Hint

\(
\begin{aligned}
&\text { (d) Energy, }\\
&\begin{aligned}
U & =\frac{1}{2} L i_0^2=\frac{1}{2} L\left(\frac{V}{R}\right)^2 \\
& =\frac{1}{2} \times 50 \times 10^{-6}\left(\frac{5}{0.5}\right)^2 \mathrm{~J} \\
& =2.5 \mathrm{~mJ}
\end{aligned}
\end{aligned}
\)
Now, after switching OFF the battery, this energy is dissipated in coil (resistance \(=0.5 \Omega\) ) and resistance ( \(10 \Omega\) ) in the ratio of their resistances ( \(H=i^2 R t\) or \(H \propto R\) ). Therefore, heat generated in the coil.
\(
2.5 \times\left(\frac{0.5}{10+0.5}\right)=0.12 \mathrm{~mJ}
\)

Hint

(b) Power, \(P=\left(i_0\right)^2 R\), i.e., \(\left(i_0\right)^2=\frac{P}{R}\) and \(\tau=\frac{L}{R}\)
Total heat produced, \(U=\frac{1}{2} L i_0^2=\frac{1}{2}(\tau R)\left(\frac{P}{R}\right)=\frac{1}{2} P \tau\)

Hint

(a) Here, area, \(A=10 \times 5=50 \mathrm{~cm}^2=50 \times 10^{-4} \mathrm{~m}^2\)
\(
\frac{d B}{d t}=0.02 \mathrm{Ts}^{-1} \text { and } R=2 \Omega
\)
\(
\mathrm{emf}, e=\frac{d \phi}{d t}=A \cdot \frac{d B}{d t}=50 \times 10^{-4} \times 0.02=10^{-4} \mathrm{~V}
\)
\(
\begin{aligned}
&\text { Power dissipated in the form of heat }\\
&=\frac{e^2}{R}=\frac{10^{-4} \times 10^{-4}}{2}=0.5 \times 10^{-8} \mathrm{~W}=5 \times 10^{-9} \mathrm{~W}=5 \mathrm{nW}
\end{aligned}
\)

Hint

(d) The magnetic field \(\otimes\) passing from the closed loop is increasing. Therefore, from Lenz’s law, induced current will produce upward magnetic field. Hence, induced current is anti-clockwise, i.e. \(I_1\) is in the direction \(a b\) and \(I_2\) is in the direction \(d c\).

Hint

(a) Since, here only half area of ring will be considered.
\(
\begin{aligned}
& \therefore \phi_B=B S=B \cdot \frac{\pi r^2}{2} \\
& \therefore \quad e=\frac{d \phi_B}{d t}=\frac{\pi r^2}{2} \cdot \frac{d B}{d t}=\frac{\pi(0.1)^2}{2} \times 0.01=\frac{\pi}{2} \times 10^{-4} \mathrm{~V} \\
& \therefore \quad i=\frac{e}{R}=\frac{\frac{\pi}{2} \times 10^{-4}}{4}=\frac{\pi}{8} \times 10^{-4}=3.9 \times 10^{-5} \mathrm{~A}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { (a) } \therefore \text { emf, }|e|=\frac{\Delta \phi}{\Delta t}=\frac{B S-0}{\Delta t}=\frac{2 \times 10^{-4} \times \pi \times(0.1)^2}{2 \times 2} \\
& \quad \Rightarrow \quad|e|=1.57 \times 10^{-6} \mathrm{~V}
\end{aligned}
\)

Hint

(b)

Due to \(v \cos 60^{\circ}, e=0\)
Due to \(v \sin 60^{\circ}, e=B\left(v \sin 60^{\circ}\right) l\)
\(
\begin{aligned}
& =\frac{\sqrt{3}}{2} B v l=\frac{\sqrt{3}}{2} \times 1 \times 0.2 \times 0.1 \\
& =17.3 \times 10^{-3} \mathrm{~V} \simeq 17 \times 10^{-3} \mathrm{~V}
\end{aligned}
\)

Hint

(c) As, the potential due to a straight rod of length \(L\) in magnetic field \(B\) is
\(
e=B L v
\)
So, for \(P Q=L / 2, e=\frac{B L v}{2}\)
Hence, current is from \(P\) to \(Q\), i.e. \(P\) is at positive potential w.r.t. \(Q\).

Hint

(a) Area of square loop, \(S=10 \mathrm{~cm} \times 10 \mathrm{~cm}\)
\(
\Rightarrow S=100 \mathrm{~cm}^2=100 \times 10^{-4} \mathrm{~m}^2=10^{-2} \mathrm{~m}^2
\)
Initial magnetic flux linked with loop,
\(
\begin{aligned}
\phi_{B_1} & =B_1 S \cos \phi=0.1 \times 10^{-2} \times \cos 45^{\circ} \\
& =\frac{0.1 \times 10^{-2} \times 1}{\sqrt{2}}=\frac{10^{-3}}{\sqrt{2}} \mathrm{~Wb}
\end{aligned}
\)
Final magnetic flux linked with loop,
\(
\phi_{B_2}=0 \mathrm{~Wb} \quad\left(\because B_2=0\right)
\)
\(
\begin{aligned}
&\text { The induced emf in the loop, }\\
&\begin{aligned}
\varepsilon & =-\frac{d \phi}{d t}=-\frac{\left(\phi_2-\phi_1\right)}{t} \\
& =-\frac{\left(0-\frac{10^{-3}}{\sqrt{2}}\right)}{0.7}=\frac{10^{-3}}{0.7 \times \sqrt{2}} \cong 10^{-3} \mathrm{~V}
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
&\text { The induced current in the loop, }\\
&I=\frac{\varepsilon}{R}=\frac{10^{-3} \mathrm{~V}}{1 \Omega}=10^{-3} \mathrm{~A}=1.0 \mathrm{~mA}
\end{aligned}
\)

Hint

(c) When the key is closed, a changing current in coil A will induce a current in coil B due to electromagnetic induction. This induced current will cause a momentary deflection in the galvanometer connected to coil B . The galvanometer will show a deflection initially when the key is closed and another deflection when the key is opened, but in the opposite direction. If the AC supply has a low frequency (e.g., 1-2 Hz), oscillations in the galvanometer may be observed.

Hint

(d)

If current \(I\) through \(A\) increases, magnetic field linked with coil \(B\) increases. Hence, anti-clockwise current induces in coil \(B\). As shown in figure, both the currents flowing in opposite directions produce repulsive effect.

Hint

\(
\begin{aligned}
&\text { (c) As, the emf induced, }\\
&\begin{aligned}
& =B_1 v l-B_2 v l=\frac{\mu_0 I}{2 \pi(x-a / 2)} v l-\frac{\mu_0 I}{2 \pi(x+a / 2)} v l \\
\Rightarrow \quad & e \propto \frac{1}{(2 x-a)(2 x+a)}
\end{aligned}
\end{aligned}
\)

Hint

(c) Circuit can be reduced as

\(
I=\frac{e}{3 R / 2}=\frac{2 v l B}{3 R} \Rightarrow I_1=I_2=\frac{I}{2}=\frac{v l B}{3 R}
\)

Hint

\(
\begin{aligned}
&\text { (c) Magnetic field at } P, B_1=\frac{\mu_0 I_2}{2 R}\\
&=\frac{4 \pi \times 10^{-7} \times 4}{2 \times 0.02 \pi}=4 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^2
\end{aligned}
\)

\(
\begin{aligned}
&\text { Magnetic field, at } Q \text {, }\\
&\begin{aligned}
B_2 & =\frac{\mu_0 I_1}{2 R}=\frac{4 \pi \times 10^{-7} \times 3}{2 \times 0.02 \pi}=3 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^2 \\
\therefore \quad B & =\sqrt{B_1^2+B^2{ }_2}=\sqrt{\left(4 \times 10^{-5}\right)^2+\left(3 \times 10^{-5}\right)^2} \\
& =5 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^2
\end{aligned}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (d) During decay of current, }\\
&i=i_0 e^{-\frac{R t}{L}}=\frac{E}{R} e^{-\frac{R t}{L}}=\frac{100}{100} e^{-\frac{100 \times 10^{-3}}{100 \times 10^{-3}}}=\left(\frac{1}{e}\right) \mathrm{A}
\end{aligned}
\)

Hint

(a) The self-inductance of a solenoid, \(L=\mu_0 n^2 S l\), where
\(\mu_0=\) permeability of air,
\(n=\) number of turns per unit length,
\(S=\) area of cross-section
and \(\quad l=\) length of the solenoid.
This depends on the geometry of the inductor such as cross-sectional area, length and number of turns and not on the material, even if it is made of a superconducting material.
\(
\therefore \quad L_1=L_2=L_3
\)

Hint

(d) Current, \(I=\frac{\text { Potential difference }}{\text { Resistance }}\)
\(\therefore \quad\) Potential difference \(=2 \times 12=24 \mathrm{~V}=B v l\)
Here, \(l=A D \sin 37^{\circ}=0.3 \times \frac{3}{5}=0.18 \mathrm{~m}\)
\(\therefore\) Velocity, \(v=\frac{24}{B l}=\frac{24}{4 \times 0.18}=\frac{100}{3} \mathrm{~ms}^{-1}\)

Hint

\(
\text { (d) Induced emf, } V_{B A}=\frac{(E \cdot 8 R)}{4}=\frac{1}{4}\left(\frac{d \phi}{d t}\right)
\)

\(
=\frac{1}{4}\left(\pi R^2\right) \cdot \frac{d B}{d t}=\frac{\pi R^2 \alpha}{4} \quad\left(\because \phi=B \cdot A \text { and } \frac{d B}{d t}=\alpha\right)
\)

Hint

(a) According to Fleming’s right hand rule, \(P\) is at higher potential and \(Q\) is at lower potential. Therefore, \(A\) is positively charged and \(B\) is negatively charged.
Also, charge, \(Q=C V=C(B v l)\)
\(
\begin{aligned}
& =10 \times 10^{-6} \times 4 \times 2 \times 1=80 \mu \mathrm{C} \\
\therefore \quad q_A & =80 \mu \mathrm{C} \text { and } q_B=-80 \mu \mathrm{C}
\end{aligned}
\)

Hint

\(
\text { (c) For rotating rod, induced emf, } v=\frac{1}{2} B l^2 \omega
\)

For part \(A O, \quad V_{O A}=V_O-V_A=\frac{1}{2} B l^2 \omega\)
For part \(O C, \quad V_{O C}=V_O-V_C=\frac{1}{2} B(3 l)^2 \omega\)
\(
\therefore V_A-V_C=V_{O C}-V_{O A}=\left(V_O-V_C\right)-\left(V_O-V_A\right)=4 B l^2 \omega
\)

Hint

\(
\begin{aligned}
&\text { (b) In } L-R \text { circuit, for growth of current, }\\
&\begin{array}{r}
i=i_0\left(1-e^{-R t / L}\right) \text { or } \frac{3}{4} i_0=i_0\left(1-e^{-t / \tau_L}\right) \text { (given) } \\
\text { (where, } \tau_L=\frac{L}{R}=\text { time constant) }
\end{array}
\end{aligned}
\)
\(
\begin{aligned}
\therefore \quad e^{-t / \tau_L} & =1-\frac{3}{4}=\frac{1}{4} \\
e^{t / \tau_L} & =4 \text { or } \frac{t}{\tau_L}=\ln 4 \\
\tau_L & =\frac{t}{\ln 4}=\frac{4}{2 \ln 2} \Rightarrow \tau_L=\frac{2}{\ln 2} \mathrm{~s}
\end{aligned}
\)

Hint

(d) As, the coil is rotated, angle \(\theta\) (angle which is normal to the coil makes with \(\mathbf{B}\) at any instant \(t\) ) changes, therefore magnetic flux \(\phi\) linked with the coil changes and hence an emf is induced in the coil. At this instant \(t\), if \(e\) is the emf induced in the coil, then
\(
e=-\frac{d \phi}{d t}=-\frac{d}{d t}(N S B \cos \omega t)
\)
where, \(N\) is number of turns in the coil.
or \(e=-N S B \frac{d}{d t}(\cos \omega t)=-N S B(-\sin \omega t) \omega[latex]\)
or \([latex]e=N S B \omega \sin \omega t \dots(i)[latex]\)
The induced emf will be maximum, when \(\sin \omega t\) is maximum.
\(
\begin{aligned}
\therefore & & e_{\max } & =e_0=N S B \omega \times 1 \\
\text { or } & & e & =e_0 \sin \omega t
\end{aligned}
\)
Therefore, \(e\) would be maximum, hence current is maximum (as \(i_0=e_0 / R\) ), when \(\theta=90^{\circ}\), i.e. normal of plane of coil is perpendicular to the field or plane of coil is parallel to magnetic field.

Hint

(a) 

For segment \(b c\), emf \(e=B v l \sin 90^{\circ}=B v(b c)\)
For segment \(a c\), emf \(a c: e=B v(a c) \sin 0^{\circ}=0\)
For segment \(a b\), emf \(a b: e=B v(a b) \sin \theta=B v(b c)\)
Since, \((b c)=(a b) \sin \theta\)

\(
\text { For whole loop, } e^{\prime}=B v l-B v l=0
\)

Hint

(c) At \(t=0\) inductor behaves as broken wire, then \(i=\frac{V}{R_2}\).

\(
\begin{aligned}
&\text { At } t=\infty \text {, inductor behaves as a conducting wire, }\\
&i=\frac{V}{R_1 R_2 /\left(R_1+R_2\right)}=\frac{V\left(R_1+R_2\right)}{R_1 R_2}
\end{aligned}
\)

Hint

(a) (i) Just before closing the switch,

As, \(i_1=0, i_2=\frac{E}{R}, i_3=\frac{E}{2 R}\),
So, \(i_2>i_3>i_1,\left[i_1=0\right]\).

(ii) After a long time closing the switch,

\(
\begin{aligned}
&\begin{array}{ll}
\text { As, } & i=\frac{E}{R} \\
\Rightarrow & i_1=E / 2 R, i_2=2 E / R, i_3=E / R
\end{array}\\
&\text { Therefore, } i_2>i_3>i_1 \text {. }
\end{aligned}
\)

Hint

(a) When the north pole of a bar magnet moves towards the coil, the induced current in the coil flows in a direction such that the coil presents its north pole to the bar magnet as shown in Fig. (a). Therefore, the induced current flows in the coil in the anti-clockwise direction.

Hint

(b) When ring enters and leaves, then the field polarity of induced emf is opposite. Also, during the stay of ring completely in the field there is no induction. These variations are correctly shown in (b).

Hint

Hint

(d) When two solenoids of inductance \(L_0[latex] are connected in series at large distance and current [latex]i\) is passed through them, then the total flux linkage, \(\phi_{\text {total }}=L_0 i+L_0 i\).
If \(L\) be the equivalent inductance of the system, then
\(
\begin{aligned}
\phi_{\text {total }} & =L i \\
L i & =L_0 i+L_0 i \quad \text { or } \quad L=2 L_0
\end{aligned}
\)
When solenoids are connected in series with one inside the other and senses of the turns coinciding, then there will be a mutual inductance \(L\) between them. In this case, the resultant induced emf in the coils is the sum of the emf’s \(e_1\) and \(e_2\) in the respective coils, i.e.
\(
\begin{aligned}
e & =e_1+e_2 \\
& =\left(-L_0 \frac{d i}{d t} \pm L_0 \frac{d i}{d t}\right)+\left(-L_0 \frac{d i}{d t} \pm L_0 \frac{d i}{d t}\right)
\end{aligned}
\)
where, \((+)\) sign is for positive coupling and \((-)\) sign for negative coupling.
But, \(e=-L \frac{d i}{d t}\)
\(
\therefore-L \frac{d i}{d t}=-L_0 \frac{d i}{d t}-L_0 \frac{d i}{d t} \pm 2 L_0 \frac{d i}{d t}
\)
\(
\text { i.e. } L=L_0+L_0+2 L_0=4 L_0 \quad \text { (for positive coupling) }
\)
When solenoids are connected in series with one inside the other with senses of the turn opposite, then there is a negative coupling.
So, \(L=L_0+L_0-2 L_0=0\)

Hint

(c) We can show the situation as

Since, loop is moving away from the wire, so the direction of current in the loop will be as shown in the figure.
Net magnetic field on the loop due to wire,
\(
B=\frac{\mu_0 i}{2 \pi}\left(\frac{1}{x}-\frac{1}{l+x}\right)=\frac{\mu_0 i l}{2 \pi x(l+x)}
\)
So, the magnitude of the emf in the loop,
\(
e=v B b=\frac{\mu_0 i l v b}{2 \pi x(l+x)}
\)

Hint

(b)

Let at any time, normal to loop makes an angle \(\theta\) with magnetic field,
\(
\begin{aligned}
\phi_B & =N B S \cos \theta=N B S \cos \omega t \\
e & =-\frac{d \phi_B}{d t}=N B S \omega \sin \omega t=e_0 \sin \omega t \\
\therefore e_{\max } & =e_0=N S B \omega=1 \times 0.01 \times \pi(5)^2 \times 10^{-4} \times \frac{80 \times 2 \pi}{60} \\
& =6.6 \times 10^{-4} \mathrm{~V}
\end{aligned}
\)
Over a long period, i.e. one time period, average of induced emf,
\(
\bar{e}=\frac{\int_0^T e d t}{T}=0
\)
and
\(
\overline{e^2}=\frac{\int_0^T e^2 d t}{T}=\frac{e_0^2}{2}=2.0 \times 10^{-7} \mathrm{~V}^2
\)

Hint

\(
\text { (c) As, } \quad E l=\frac{d \phi_B}{d t}
\)
\(
\text { or } \quad E(2 \pi R)=\pi R^2 \cdot \frac{d B}{d t}=\pi R^2(8 t)
\)
\(
\begin{aligned}
& E=4 R t \\
& F=q E=4 q R t \text { (tangential) }\\
& F=4 q R^2 t \\
& f=(\mu m g R)
\end{aligned}
\)
When \(F>f\), ring will start rotating.
At
\(
\begin{gathered}
t=2 \mathrm{~s} \\
8 q R^3=\mu m g R
\end{gathered}
\)
\(\therefore\) Coefficient of friction, \(\mu=\frac{8 q R}{m g}\)

Hint

(b)

Draw a concentric circle of radius \(r\). The induced electric field \(E\) at any point on the circle is equal to that at \(P\).
For this circle, induced emf,
\(
e=\oint \mathbf{E} \cdot d \mathbf{l}=\left|\frac{d \phi}{d t}\right|=S\left|\frac{d B}{d t}\right|
\)
\(
e=E \int d l=\pi a^2\left|\frac{d B}{d t}\right| \quad \text { (but } \oint d l=2 \pi r \text { ) }
\)
\(
\begin{array}{ll}
\therefore & E \times 2 \pi r=\pi a^2\left|\frac{d B}{d t}\right| \\
\therefore & E=\frac{a^2}{2 r}\left|\frac{d B}{d t}\right| \\
\Rightarrow & E \propto \frac{1}{r}
\end{array}
\)

Hint

(d) When switch \(S\) is closed, then magnetic field lines passing through \(Q\) increases in the direction from right to left.
So, according to Lenz’s law, induced current in \(Q\), i.e. \(I_{Q_1}\) will flow in such a direction, so that the magnetic field lines due to \(I_{Q_1}\) passes from left to right through \(Q\). This is possible when \(I_{Q_1}\) flows in anti-clockwise direction as seen by \(E\). Opposite is the case when switch \(S\) is opened, i.e. \(I_{Q_2}\) will be clockwise as seen by \(E\).

Hint

(a) As the north pole approaches, a north pole is developed at the face, i.e. the current flows anti-clockwise. Finally, when it completes the oscillation, no emf is present.
Now, south pole approaches the other side, i.e. RHS, the current flows clockwise to repel the south pole. This means the current is anti-clockwise at the LHS of the coil as before. The break occurs, when the pendulum is at the extreme and momentarily stationary position.

Hint

(b) Lenz’s Law is based on the principle of conservation of energy.
Explanation:
Lenz’s Law states that the direction of the induced current in a conductor due to a changing magnetic field is such that it creates a magnetic field that opposes the initial changing magnetic field. This opposing action is crucial for energy conservation.

If the induced current reinforced the change in magnetic flux, it would lead to a runaway scenario where energy is created without any work being done, violating the law of conservation of energy.

Lenz’s Law ensures that the work done against the opposing force of the induced magnetic field is converted into electrical energy, thereby conserving the total energy of the system.

Hint

(a) When a metallic surface is moved in and out in magnetic field, then magnetic flux linked with metallic surface is changed continuously, hence according to Faraday’s law of electromagnetic induction, an induced emf is produced. Since, metallic plate behaves as a closed path, therefore an induced current (eddy current) starts flowing in the metallic surface.

Hint

(a) Coil \(A\) carries a steady current with increase in temperature, its resistance increases and so current is decreasing at a constant rate, this induces an emf in coil \(B\) which opposes this change, i.e. current in coil \(B\) is in same direction of \(A\), therefore they attract to each other.

Hint

(a) Given, \(n=500\),
\(
\begin{aligned}
& L=50 \mathrm{mH}=50 \times 10^{-3} \mathrm{H} \\
& i=8 \mathrm{~mA}=8 \times 10^{-3} \mathrm{~A}
\end{aligned}
\)
The magnetic flux linked with the coil,
\(
\begin{aligned}
\phi & =L i=50 \times 10^{-3} \times 8 \times 10^{-3} \\
& =400 \times 10^{-6}=4 \times 10^{-4} \mathrm{~Wb}
\end{aligned}
\)

Hint

(a) The flux is given by
\(
\begin{aligned}
& \phi=\mathbf{B} \cdot \mathbf{A} \\
& \phi=|\mathbf{B}||\mathbf{A}| \cos \omega t
\end{aligned}
\)
Also, induced emf, \(e=-\frac{d \phi}{d t}=-\frac{d[|\mathbf{B}||\mathbf{A}| \cos \omega t]}{d t}\)
\(
\begin{aligned}
& =-|\mathbf{B}||\mathbf{A}| \frac{d(\cos \omega t)}{d t}=|\mathbf{B}||\mathbf{A}| \sin \omega t \cdot \omega \\
\Rightarrow \quad & e=B A \omega \cos (\omega t-\pi / 2)
\end{aligned}
\)

Hint

(b) Given, \(B=40 \mathrm{mT}=40 \times 10^{-3} \mathrm{~T}\)
Rate of shrinking of area \(=0.5 \mathrm{~m}^2 \mathrm{~s}^{-1}\)
\(
\Rightarrow \quad \frac{d A}{d t}=0.5 \mathrm{~m}^2 \mathrm{~s}^{-1}
\)
\(
\text { As, } \quad \phi=B A \cos \theta=B A \quad\left(\because \cos \theta=\cos 0^{\circ}=1\right)
\)
On differentiating with w.r.t. to \(t\),
\(
\frac{d \phi}{d t}=B \cdot \frac{d A}{d t}=e
\)
\(
\begin{aligned}
e & =B \cdot \frac{d A}{d t}=40 \times 10^{-3} \times 0.5 \\
& =20 \times 10^{-3} \mathrm{~V} \\
& =20 \mathrm{mV}
\end{aligned}
\)

Hint

(a) When a changing magnetic field induces currents in a nearby metal body, these currents are called eddy currents.

Explanation: Eddy currents are circular currents that form within a conductor when exposed to a changing magnetic field, essentially swirling currents within the metal.

Why other options are incorrect:
Flux currents:
While related to magnetic flux, “flux currents” is not a standard term for the currents induced by a changing magnetic field.

Alternating currents:
AC refers to the direction of current periodically reversing. Eddy currents can be AC or DC depending on the nature of the changing magnetic field, but the term “alternating current” describes the current itself, not the mechanism of its creation.

Leaking currents:
Leakage currents refer to unintended currents that escape from a circuit due to insulation faults. They are not related to the phenomenon of eddy currents.

Hint

(b) Consider a rod of 10 cm length is moving perpendicular to uniform magnetic field as shown below

The emf induced in the rod, \(e=v B l\) Differentiating with respect to time \(t\), we get
\(
\begin{aligned}
& \text { Rate of increase of induced emf }=\frac{d e}{d t} \\
& \qquad \begin{aligned}
= & \frac{d v}{d t} B l=a B l \\
= & 5 \times 5 \times 10^{-4} \times 10 \times 10^{-2} \\
= & 250 \times 10^{-6} \\
= & 2.5 \times 10^{-4} \mathrm{Vs}^{-1}
\end{aligned}
\end{aligned}
\)

Hint

(b) Induced emf for both the loops will be given as
\(
e=\frac{-d \phi}{d t}=B \omega \sin \omega t
\)
Since, \(B\) and \(\omega\) are same for both the cases, hence induced emf will be same.
However, induced current is given by
\(
I=\frac{e}{R}
\)
So, it will depend upon the value of resistance, lower the resistance higher will be the current. Therefore, induced current is different in both the loops.

Hint

(b) Given, length of conductor, \(l=0.1 \mathrm{~m}\)
Mangetic field, \(B=0.1 \mathrm{~T}\)
Velocity of conductor, \(v=15 \mathrm{~ms}^{-1}\)
The angle between \(v\) and \(B\) is \(90^{\circ}\).
When \(v\) and \(B\) are mutually perpendicular, then emf (induced) is given by
\(
\begin{aligned}
e & =v B l=15 \times 0.1 \times 0.1 \\
& =\frac{15}{100}=0.15 \mathrm{~V}
\end{aligned}
\)

Hint

(c) Applying KVL in the circuit, we can write

\(
\begin{aligned}
&6-L \frac{d I}{d t}-I r=0\\
&\text { At initial stage, } \quad I=0\\
&\Rightarrow \quad 6=L \frac{d I}{d t} \text { or } \frac{d I}{d t}=\frac{6}{L}=\frac{6}{2}=3 \mathrm{As}^{-1}
\end{aligned}
\)

Hint

(d) As magnetic field through circle is constant because current through straight wire is constant. Magnetic flux through circle is not changing with time.
\(
\begin{gathered}
\frac{d \phi_B}{d t}=0 \\
\Rightarrow \quad|e|=\left|\frac{d \phi_B}{d t}\right|=0 \Rightarrow i=\frac{e}{R}=\frac{0}{R}=0
\end{gathered}
\)
Thus, no current flow in the circle.

Hint

(c) We know that, \(e=N B A \omega \sin \omega t\)
where, \(N=\) number of loops \(=1\)
\(
\begin{aligned}
B & =\frac{\mu_0 I}{2 b} \mathrm{~N} / \mathrm{A}-\mathrm{m} \Rightarrow A=\pi a^2 \mathrm{~m}^2 \\
e & =\frac{\mu_0 I}{2 b}\left(\pi a^2\right) \omega \sin \omega t \\
& =\frac{\pi a^2 \mu_0 I}{2 b} \cdot \omega \sin \omega t
\end{aligned}
\)

Hint

(c) Consider the diagram, where a conductor \(A B\) of length \(l\) moving perpendicular to an inward magnetic field \(B\).

\(
\begin{aligned}
&\text { Value of induced emf, }\\
&\left|E_{\text {ind }}\right|=|l(\mathbf{v} \times \mathbf{B})|
\end{aligned}
\)
\(
\Rightarrow \quad=l v B \sin 90^{\circ} \quad(\because v \perp B)
\)
\(
E_{\text {ind }}=v B l \dots(i)
\)
Given, \(v=7 \mathrm{~ms}^{-1}, l=0.4 \mathrm{~m}\) and \(B=0.9 \mathrm{Wbm}^{-2}\)
Now, from Eq. (i), we get
\(
E_{\text {ind }}=7(0.9)(0.4) \mathrm{V}=2.52 \mathrm{~V}
\)

Hint

(a) Consider the inductor of inductance \(L\) and the current flowing through the inductor is \(i\).
Now, we can write \(\phi=L i\) where, \(\phi\) is magnetic flux linked with the inductor.
\(
\begin{array}{ll}
\Rightarrow & \frac{d \phi}{d t}=L \frac{d i}{d t} \\
\text { Given, } & L=40 \mathrm{mH}
\end{array}
\)
\(
d t=\text { change in time }=t_2-t_1=4 \mathrm{~ms}=\Delta t
\)
and \(d i=i_2-i_1=11-1=10 \mathrm{~A}=\Delta i\)
So, \(\quad \frac{d \phi}{d t}=\frac{\Delta \phi}{\Delta t}=L \frac{\Delta i}{\Delta t}=40 \times \frac{10}{4}\)
\(
=10 \times 10=100 \dots(i)
\)
According to Faraday’s law of electromagnetic induction, Emf induced, \(e=-\frac{d \phi}{d t}\)
\(
\Rightarrow \quad|e|=\frac{\Delta \phi}{\Delta t}=100 \mathrm{~V} \quad \text { [from Eq. (i)] }
\)

Hint

\(
\begin{aligned}
& \text { (b) } \because \text { Induced emf, } e=-L \frac{d i}{d t} \\
& \text { Given, } \quad L=10 \mathrm{H}, \Delta i=(9-4) \mathrm{A}=5 \mathrm{~A}, d t=0.2 \mathrm{~s} \\
& \text { emf, } \quad e=10 \times \frac{5}{0.2}=250 \mathrm{~V}
\end{aligned}
\)

Hint

(c) As, \(e=B v l\)
Here, \(B=10^{-3} \mathrm{~Wb} / \mathrm{m}^3\)
\(
\begin{aligned}
& \quad v=2 \mathrm{~m} / \mathrm{s} \text { and } l=5 \mathrm{~cm}=0.05 \mathrm{~m} \\
& \therefore \mathrm{emf}, \quad e=\left(10^{-3}\right)(2)(0.05)=1 \times 10^{-4} \mathrm{~V}
\end{aligned}
\)

Hint

(c) As the current in coil \(A\) changes, the magnetic flux linked with coil \(B\) changes. Direction of flux in \(B\) will be downwards, hence current induces in coil \(B\) is in anti-clockwise direction according to Lenz’s law.
The direction of induced current in coil \(B\) is opposite to that of the direction of current in coil \(A\) as shown in figure.

Due to currents in opposite directions in the near by sides of the coil, the coil \(B\) is repelled.

Hint

(c) Given, \(n=100\) turns, \(A=(0.1 \times 0.05) \mathrm{m}^2\)
\(
B_1=0.1 \mathrm{~T}, B_2=0.05 \mathrm{~T}, d t=0.05 \mathrm{~s}
\)
We know that,
\(
e=\left|\frac{-d \phi}{d t}\right|=\left|\frac{d}{d t}(n B A \cos \theta)\right|=\left|\frac{n A d B \cos \theta}{d t}\right|
\)
Here, \(\quad \theta=0^{\circ}\)
\(
\begin{array}{ll}
\therefore & e=\left|\frac{n A d B}{d t}\right|=\left|\frac{100 \times 0.1 \times 0.05 \times(0.1-0.05)}{0.05}\right| \\
\Rightarrow & e=0.5 \mathrm{~V}
\end{array}
\)

Hint

(a) Electromagnetic induction is not used in room heater because it works on the principle of heating effect of electric current.

Hint

(d) Given, \(M=0.05 \mathrm{H}[latex] and [latex]I=I_0 \sin \omega t\)
\(
\begin{aligned}
& \therefore \quad \frac{d I}{d t}=I_0 \cos \omega t(\omega) \\
& \Rightarrow\left(\frac{d I}{d t}\right)_{\max }=I_0(\omega) \times 1=1 \times 100 \pi \mathrm{~A} / \mathrm{s}
\end{aligned}
\)
So, maximum emf,
\(
\begin{aligned}
e_{\max } & =M\left(\frac{d I}{d t}\right)_{\max } \\
& =0.05 \times 100 \pi \\
& =5 \pi \mathrm{~V}
\end{aligned}
\)

Hint

(d) For inductor, as we know induced voltage,
(for \(t=0\) to \(t=T / 2\) )
\(
V=L \frac{d I}{d t}=L \frac{d}{d t}\left(\frac{2 I_0 t}{T}\right)=\text { constant }
\)
(for \(t=T / 2\) to \(t=T\)
\(
V=\frac{L d I}{d t}\left(\frac{-2 I_0 t}{T}\right)=- \text { constant }
\)
Therefore, correct variation will be represented by graph (d).

Hint

(d) Induced emf,
\(
e=-L \frac{d i}{d t}
\)
During 0 to \(\frac{T}{4}, \frac{d i}{d t}=\) constant and positive, so \(e=\) negative.
For \(\frac{T}{4}\) to \(\frac{T}{2}, \frac{d i}{d t}=0\), so \(e=0\).
For \(T / 2\) to \(\frac{3 T}{4}, \frac{d i}{d t}=\) constant and negative, so \(e=\) positive.
This condition is satisfied only by graph in option (d).

Hint

\(
\begin{aligned}
& \text { (a) Given, } B=0.15 \mathrm{~T}, l=50 \mathrm{~cm}=50 \times 10^{-2} \mathrm{~m} \\
& \text { and } v=2 \mathrm{~ms}^{-1} \\
& \therefore \quad \text { emf, } e=B v l \\
& \qquad e=0.15 \times 2 \times 50 \times 10^{-2}=0.15 \mathrm{~V}
\end{aligned}
\)
\(
\begin{aligned}
& \text { Current, } i=\frac{e}{R}=\frac{0.15}{3}=5 \times 10^{-2} \mathrm{~A} \\
& \text { Force, } F=\text { Bil }=0.15 \times 5 \times 10^{-2} \times 50 \times 10^{-2} \\
& \\
& =3.75 \times 10^{-3} \mathrm{~N}
\end{aligned}
\)

Hint

(a) Induced emf, \(e=2 \mathrm{~V}\)
\(
i_1=8 \mathrm{~A}, i_2=6 \mathrm{~A} \text { and } \Delta t=2 \times 10^{-3} \mathrm{~s}
\)
Coefficient of self-induction,
\(
\begin{aligned}
L & =\frac{e}{\Delta i / \Delta t}=\frac{-2}{(6-8) / 2 \times 10^{-3}} \\
& =\frac{-2 \times 2 \times 10^{-3}}{-2} \\
& =2 \times 10^{-3} \mathrm{H}
\end{aligned}
\)

Hint

(c) Rate of doing work \(=\frac{W}{t}=P\)
But, \(\quad P=F v\)
Also, \(\quad F=B i l=B\left(\frac{B v l}{R}\right) l\left[\because\right.\) induced current, \(\left.i=\frac{B v l}{R}\right]\)
\(\therefore\) Power, \(P=B\left(\frac{B v l}{R}\right) l v=\frac{B^2 v^2 l^2}{R}\)
\(
=\frac{(0.5)^2 \times(2)^2 \times(1)^2}{6}=\frac{1}{6} \mathrm{~W}
\)

Hint

(a) Given, \(\phi=10 \mathrm{~Wb}, \phi_2=60 \mathrm{~Wb}\) and \(R=100 \Omega\)
The amount of induced charge is given by
\(
\begin{aligned}
q & =\frac{1}{R} \Delta \phi=\frac{1}{100}(60-10) \\
\therefore \text { Charge, } q & =\frac{50}{100}=0.5 \mathrm{C}
\end{aligned}
\)

Hint

(a) Self-inductance, \(L=\frac{\mu_0 N^2 \pi r^2}{l}\)
Given, \(r_1 / r_2=l_1 / l_2=1 / 2\)
\(
\begin{aligned}
& \Rightarrow \quad \frac{L_1}{L_2}=\left(\frac{r_1}{r_2}\right)^2\left(\frac{l_2}{l_1}\right)=\left(\frac{1}{2}\right)^2 \times\left(\frac{2}{1}\right) \\
& \Rightarrow \quad \frac{L_1}{L_2}=\frac{1}{2}=1: 2
\end{aligned}
\)

Hint

(b) For two such coils connected oppositely in series,
\(
\begin{aligned}
L_{\mathrm{eq}} & =L_1+L_2-2 M \\
L_1 & =L_2=L \text { and } M=0 \\
L_{\mathrm{eq}} & =L+L-0=2 L
\end{aligned}
\)