NEET Electromagnetic Induction MCQs & PYQs

Practice NEET Electromagnetic Induction Questions

Question 81:

easy

22. A big circular coil of 100 turns and average radius $10 m$ is rotating about its horizontal diameter at $2 rad s^{-1}$. If the vertical component of earth’s magnetic field at that place is $2 \times 10^{-5} T$ and electrical resistance of the coil is $12.56 \Omega$, then the maximum induced current in the coil will be : (2022)

The peak induced emf is given by $e_0 = N B A \omega = N B (\pi r^2) \omega$.
Substituting the values: $e_0 = 100 \times (2 \times 10^{-5}) \times (\pi \times 10^2) \times 2 = 4\pi \approx 12.56 V$.
The maximum induced current is $I_0 = \frac{e_0}{R} = \frac{12.56}{12.56} = 1 A$.

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Question 82:

easy

24. A metallic rod of mass per unit length $0.5 kg m^{-1}$ is lying horizontally on a smooth inclined plane which makes an angle of $30^\circ$ with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction $0.25 T$ is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is (2018)

For the rod to remain stationary on the incline, the forces along the plane must balance: $m g \sin\theta = I l B \cos\theta$.
Rearranging gives $I = \left(\frac{m}{l}\right) \frac{g \tan\theta}{B}$.
With $\frac{m}{l} = 0.5 kg m^{-1}$, $g = 9.8 m s^{-2}$, $\theta = 30^\circ$, and $B = 0.25 T$: $I = \frac{0.5 \times 9.8 \times \tan(30^\circ)}{0.25} = \frac{4.9}{0.25 \sqrt{3}} \approx 11.32 A$.

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Question 83:

easy

27. A rectangular, a square, a circular and an elliptical loop, all in the (x-y) plane, are moving out of a uniform magnetic field with a constant velocity, $\vec{v} = v\hat{i}$. The magnetic field is directed along the negative z axis direction. The induced emf, during the passage of these loops, out of the field region, will not remain constant for: (2009)

The magnitude of induced emf is $e = B l_{eff} v$, where $l_{eff}$ is the length of the conductor cutting the field lines.
For rectangular and square loops, $l_{eff}$ is constant as they exit the field, resulting in a constant emf.
For circular and elliptical loops, $l_{eff}$ continuously varies with time, so the induced emf does not remain constant.

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Question 84:

easy

28. A straight line conductor of length $0.4 m$ is moved with a speed of $7 m/s$ perpendicular to a magnetic field of intensity $0.9 Wb/m^2$. The induced e.m.f. across the conductor is (1995)

The motional e.m.f. induced across a straight conductor moving perpendicular to a magnetic field is $e = B v l$.
Given $B = 0.9 Wb/m^2$, $v = 7 m/s$, and $l = 0.4 m$.
$e = 0.9 \times 7 \times 0.4 = 2.52 V$.

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Question 85:

easy

45. Two coil have a mutual inductance $0.005 H$. The current changes in first coil according to equation $I = I_0 \sin \omega t$ where $I_0 = 2 A$ and $\omega = 100pi rad/sec$. The maximum value of emf in second coil is: (1998)

Induced emf in the second coil is $e = -M \frac{dI}{dt} = -M \frac{d}{dt}(I_0 \sin \omega t) = -M I_0 \omega \cos \omega t$.
The maximum (peak) value of induced emf is $e_{max} = M I_0 \omega$.
$e_{max} = 0.005 \times 2 \times 100\pi = 0.01 \times 100\pi = \pi V$.

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Question 86:

easy

32. A long solenoid has 1000 turns. When a current of $4 A$ flows through it, the magnetic flux linked with each turn of the solenoid is $4 \times 10^{-3} Wb$. The self inductance of the solenoid is: (2016 – I)

Total magnetic flux linked with the solenoid is $N \Phi = L I$.
Given $N = 1000$, $\Phi = 4 \times 10^{-3} Wb$, and $I = 4 A$.
$L = \frac{N \Phi}{I} = \frac{1000 \times 4 \times 10^{-3}}{4} = 1 H$.

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Question 87:

easy

33. A long solenoid has 500 turns. When a current of 2 ampere is passed through it, the resulting magnetic flux linked with each turn of the solenoid is $4 \times 10^{-3} Tm^2$. The self inductance of the solenoid is: (2008)

Total flux linked is given by $N \Phi = L I$.
Here, $N = 500$, $\Phi = 4 \times 10^{-3} Wb$ (or $T m^2$), and $I = 2 A$.
$L = \frac{N \Phi}{I} = \frac{500 \times 4 \times 10^{-3}}{2} = 1.0 henry$.

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Question 88:

easy

34. If N is the number of turns in a coil, the value of self inductance varies as (1993)

Self-inductance of a coil/solenoid is given by $L = \frac{\mu_0 N^2 A}{l}$.
This relation clearly shows that self-inductance is directly proportional to the square of the number of turns.
Therefore, $L \propto N^2$.

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Question 89:

easy

35. What is the self-inductance of a coil which produces $5 V$ when the current changes from 3 ampere to 2 ampere in one millisecond? (1993)

Induced emf is given by $e = L \left|\frac{\Delta I}{\Delta t}\right|$.
Here $e = 5 V$, $|\Delta I| = |2 - 3| = 1 A$, and $\Delta t = 1 ms = 10^{-3} s$.
$L = \frac{e}{|\Delta I / \Delta t|} = \frac{5}{1 / 10^{-3}} = 5 \times 10^{-3} H = 5 mili-henry$.

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Question 90:

easy

36. If the number of turns per unit length of a coil of solenoid is doubled, the self-inductance of the solenoid will: (1991)

Self-inductance of a solenoid is $L = \mu_0 n^2 A l$, where $n$ is the number of turns per unit length.
Thus, self-inductance is directly proportional to the square of turns per unit length ($L \propto n^2$).
When $n$ is doubled, $L' = (2n)^2 = 4L$, so it becomes four times.

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