Electromagnetic Induction — NEET Physics PYQ MCQs with Solutions

Solution: Magnetic flux, $\phi = \vec{B}\cdot\vec{A} = BA\cos\theta = \dfrac{1}{\pi}\times\pi(0.2)^2\times\cos60^\circ = 0.04\times0.5 = 0.02\,\text{Wb}$.

Solution: The induced emf $V$ acts as the potential difference driving the charge once around the loop. Work done in moving a charge $q$ through a potential difference $V$ is given by $W = qV$.

Solution: From Faraday's law, emf induced in the circuit is $e = \dfrac{\Delta\phi}{\Delta t}$. If $R$ is the resistance of the circuit, current $i = \dfrac{e}{R} = \dfrac{\Delta\phi}{R\,\Delta t}$. Thus, charge passing through the circuit, $q = i\,\Delta t = \dfrac{\Delta\phi}{R\,\Delta t}\times\Delta t = \dfrac{\Delta\phi}{R}$.

Solution: Electric heaters work on the Joule heating effect of current through a resistor, not on eddy currents. Magnetic braking, electromagnets and induction furnaces all rely on eddy currents induced in bulk conductors by a changing flux.

Solution: As the loop rotates, the induced emf varies as $e=e_0\sin\omega t$, which changes sign (direction) twice in every complete revolution — once each time the plane of the loop becomes parallel to $B$.

Solution: $e=Blv\sin\theta$, with $\theta=90^{\circ}$ since $\vec B\perp\vec v$. So $e=0.9\times0.4\times7\times\sin90^{\circ}=2.52\text{ V}$.

Solution: Eddy currents are induced currents that circulate within the body of a bulk conductor when the magnetic flux linked with it changes with time, i.e. when the conductor is in a varying (not steady) magnetic field.

Solution: Inserting an iron core increases the inductance $L$ of the coil, which increases the inductive reactance $X_L=\omega L$. Since the circuit current is $I=e/X_L$, the current decreases, so the bulb's glow decreases.

Solution: $L=\dfrac{N\phi}{I}=\dfrac{1000\times4\times10^{-3}}{4}=1\text{ H}$.

Solution: Net flux linkage, $\phi_{net}=N\phi=500\times4\times10^{-3}=2\text{ Wb}$. Self-inductance $L=\dfrac{\phi_{net}}{i}=\dfrac{2}{2}=1\text{ H}$.

Solution: $e=L\dfrac{di}{dt} \Rightarrow 220=L\times\dfrac{10}{0.5} \Rightarrow L=\dfrac{220\times0.5}{10}=11\,\text{H}$.

Solution: Magnetic flux $\phi=BA$ and the field at the centre of a circular coil is $B=\dfrac{\mu_0Ni}{2R}$. The total flux linkage is $N\phi=\dfrac{\mu_0N^2iA}{2R}$, so $L=\dfrac{N\phi}{i}=\dfrac{\mu_0N^2A}{2R}$. Hence $L\propto N^2$.

Solution: $|e|=L\dfrac{di}{dt} \Rightarrow L=\dfrac{|e|}{di/dt}=\dfrac{5}{1/(1\times10^{-3})}=5\times10^{-3}\,\text{H}=5\,\text{mH}$.

Solution: $E=\dfrac{1}{2}Li^2=\dfrac{1}{2}\times(100\times10^{-3})\times1^2=0.05\,\text{J}$.

Solution: For a long solenoid, $L=\mu_0n^2Al$, where $n$ is the number of turns per unit length. So $L\propto n^2$; doubling $n$ makes $L$ four times its original value.

Solution: The induced emf is $e=-L\dfrac{di}{dt}$, and the work done in establishing a current $i_0$ is $W=\displaystyle\int_0^{i_0}Li\,di=\dfrac{1}{2}Li_0^2$. This work is stored as energy in the inductor's magnetic field.

Solution: The energy stored in an inductor carrying current $i_0$ is $W=\dfrac{1}{2}Li_0^2$, and this energy resides in the magnetic field set up by the current.

Solution: When a coil of $N$ turns and area $A$ rotates in an external magnetic field $B$, the flux linked is $\phi = NBA\cos\omega t$, so induced emf $e = -\dfrac{d\phi}{dt} = NBA\omega\sin\omega t$, with maximum current $i_0 = \dfrac{e_0}{R} = \dfrac{NBA\omega}{R}$. Given $N=1$, $B=10^{-2}\,\text{T}$, $A=\pi(0.3)^2\,\text{m}^2$, $R=\pi^2\,\Omega$, $f=\dfrac{200}{60}\,\text{s}^{-1}$, $\omega=2\pi f = 2\pi\times\dfrac{200}{60}$. So $i_0 = \dfrac{10^{-2}\times\pi(0.3)^2\times2\pi\times\frac{200}{60}}{\pi^2} = 6\times10^{-3}\,\text{A} = 6\,\text{mA}$.

Solution: The near and far sides of the square frame (at distances $x-\frac{a}{2}$ and $x+\frac{a}{2}$ from the wire) develop motional emfs $e_1=\dfrac{\mu_0IaV}{2\pi(x-a/2)}$ and $e_2=\dfrac{\mu_0IaV}{2\pi(x+a/2)}$, which oppose each other. Net emf $e=e_1-e_2\propto\dfrac{1}{x-a/2}-\dfrac{1}{x+a/2}=\dfrac{a}{x^2-a^2/4}=\dfrac{4a}{(2x-a)(2x+a)}$, so $e\propto\dfrac{1}{(2x-a)(2x+a)}$.

Solution: Charge $q=\dfrac{\Delta\phi}{R}=\dfrac{N_2\mu_0n_1\Delta I\,\pi r^2}{R}$, with $n_1=2\times10^4\ \text{m}^{-1}$, $N_2=100$, $r=0.01\text{ m}$, $\Delta I=4\text{ A}$, $R=10\pi^2\ \Omega$. $q=\dfrac{100\times4\pi\times10^{-7}\times2\times10^4\times4\times\pi\times10^{-4}}{10\pi^2}=32\times10^{-6}\text{ C}=32\ \mu\text{C}$.

Solution: Magnetic flux linked with coil, $\phi = (5t^2+3t+16)\,\text{Wb}$. Induced emf, $e = \dfrac{d\phi}{dt} = 10t+3$. At $t=3\,\text{s}$, $e_3 = 10(3)+3 = 33\,\text{V}$. At $t=4\,\text{s}$, $e_4 = 10(4)+3 = 43\,\text{V}$. Induced emf in the coil at (during) the fourth second $= e_4-e_3 = 43-33 = 10\,\text{V}$.

Solution: Given, $A=0.05\,\text{m}^2$, $B=5\times10^{-5}\,\text{T}$, $N=800$. Flux linked with the coil, $\phi = N(\vec{B}\cdot\vec{A}) = NBA\cos\theta$, where $\theta$ is the angle between $\vec{B}$ and $\vec{A}$. Emf induced when the coil is rotated from $\theta_1=0^\circ$ to $\theta_2=90^\circ$: $e = \dfrac{NBA}{\Delta t}(\cos\theta_2-\cos\theta_1)$. With $\Delta t = 0.1\,\text{s}$: $e = \dfrac{800\times5\times10^{-5}\times0.05\times[\cos90^\circ-\cos0^\circ]}{0.1} = 2000\times10^{-5} = 0.02\,\text{V}$.

Solution: Induced emf is $|e| = \dfrac{d\phi}{dt}$. For loop 1 (radius $R>r$), it encloses the entire field region of area $\pi r^2$, so $\phi = \pi r^2 B$ and $e_1 = \dfrac{d\phi}{dt} = \pi r^2 \dfrac{dB}{dt}$. Loop 2 lies entirely outside the field region, so the flux linked with it is always zero, giving $e_2 = 0$.

Solution: Induced emf in a coil is given by $E = \dfrac{d\phi}{dt}$. Given $\phi = 50t^2+4$ and $R=400\,\Omega$. So $E = \dfrac{d\phi}{dt}\Big|_{t=2} = 100t\Big|_{t=2} = 200\,\text{V}$. Current in the coil, $I = \dfrac{E}{R} = \dfrac{200}{400} = 0.5\,\text{A}$.

Solution: Magnetic flux linked with the field $\vec{B}$ and area $\vec{A}$ is $\phi = \vec{B}\cdot\vec{A} = BA = B\pi r^2$ (since $\theta=0$). Induced emf, $|e| = \left|\dfrac{d\phi}{dt}\right| = B(2\pi r)\dfrac{dr}{dt} = 0.025\times2\pi\times0.02\times0.001 = \pi\times10^{-6}\,\text{V} = \pi\,\mu\text{V}$.

Solution: For the rectangular and square loops, the leading edge is a straight side of fixed length, so the area coming out per second (and hence the induced emf) stays constant while it exits the field. For the circular and elliptical loops, the length of the boundary crossing the field edge keeps changing as the loop exits, so the area swept out per second is not constant — the induced emf during the passage of these loops out of the field region will not remain constant.

Solution: Magnetic field, $B=0.04\,\text{T}$, and rate of change of radius, $\dfrac{dr}{dt} = 2\,\text{mm}\,\text{s}^{-1} = 2\times10^{-3}\,\text{m/s}$. Induced emf, $e = \dfrac{d\phi}{dt} = B\dfrac{dA}{dt} = B\dfrac{d(\pi r^2)}{dt} = B(2\pi r)\dfrac{dr}{dt}$. At $r=2\,\text{cm}=2\times10^{-2}\,\text{m}$: $e = 0.04\times2\pi\times2\times10^{-2}\times2\times10^{-3} = 3.2\pi\times10^{-6}\,\text{V} = 3.2\pi\,\mu\text{V}$.

Solution: Total charge induced in a conducting loop is $q = \int i\,dt$. Since $i=\dfrac{e}{R}$, $q = \displaystyle\int\dfrac{e}{R}\,dt = \dfrac{1}{R}\int e\,dt = \dfrac{1}{R}\int d\phi = \dfrac{\Delta\phi}{R}$. Hence the total charge depends only on the resistance of the loop and the total change in magnetic flux, not on how fast the flux changes.

Solution: Given, $N=20$, $A=25\,\text{cm}^2=25\times10^{-4}\,\text{m}^2$, $\dfrac{dB}{dt}=1000\,\text{T/s}$, $R=100\,\Omega$. Induced current, $i = \dfrac{e}{R} = \dfrac{NA\frac{dB}{dt}}{R} = \dfrac{20\times25\times10^{-4}\times1000}{100} = \dfrac{50}{100} = 0.5\,\text{A}$.

Solution: Emf induced due to change in magnetic flux, $e = \dfrac{d\phi}{dt} = \dfrac{\phi_2-\phi_1}{dt}$. When magnetic field is perpendicular to the coil, $\phi_1 = NBA$; when coil is removed, $\phi_2=0$. So $e = \dfrac{0-NBA}{dt} \Rightarrow dt = \dfrac{NBA}{e} = \dfrac{50\times2\times10^{-2}\times100\times10^{-4}}{0.1} = 0.1\,\text{s}$.