JEE Physics Electromagnetic Induction MCQs Set 05

Answer: (b) \(25\ \text{V}\)

Explanation:
1. Motional emf \(= (\vec v\times\vec B)\cdot\vec L\).
2. \(\vec v\times\vec B = \hat i\times(3\hat i + 4\hat j + 5\hat k) = 4\hat k - 5\hat j\).
3. \(\vec L = 5\hat j\), so emf \(= -25\ \text{V}\), and its magnitude is \(25\ \text{V}\).

Teacher's Note:
1. Only the field component perpendicular to both the velocity and the rod contributes.
2. Here that is \(B_z = 5\ \text{T}\), giving \(5\times1\times5 = 25\ \text{V}\).

Answer: (b) \(5\ \text{A}\)

Explanation:
1. Induced emf \(= NA\frac{dB}{dt} = 10\times10^{-3}\times10^4 = 100\ \text{V}\).
2. \(I = \frac{100}{20} = 5\ \text{A}\).

Teacher's Note:
1. Convert the area from square centimetres to square metres: \(10\ \text{cm}^2 = 10^{-3}\ \text{m}^2\).
2. Multiply by the number of turns for the total flux linkage.

Answer: (a) clockwise in A, anticlockwise in B

Explanation:
1. By the right-hand rule, the wire's field points out of the page on its left side (through A) and into the page on its right side (through B).
2. The field through A (out of the page) increases, so the induced current in A makes a field into the page: clockwise.
3. The field through B (into the page) increases, so the induced current in B makes a field out of the page: anticlockwise.

Teacher's Note:
1. Lenz's law: the induced current opposes the change in flux.
2. Use the right-hand grip rule for the wire's field on each side.

Answer: (b) remain the same

Explanation:
1. The motional emf depends on the effective length, which is the distance between the end points of the moving wire (the chord).
2. The semicircular wire has the same end points as the straight wire, so the emf \(Blv\) is unchanged.
3. The circuit resistance is unchanged, so the current is the same.

Teacher's Note:
1. Only the component of the wire perpendicular to its velocity and to the field matters, found from the straight line joining its ends.
2. The shape of the wire between the end points does not matter.

Answer: (d) \(5\ \text{mH}\)

Explanation:
1. \(|\varepsilon| = L\left|\frac{dI}{dt}\right|\), with \(\left|\frac{dI}{dt}\right| = \frac{1}{10^{-3}} = 10^3\ \text{A/s}\).
2. \(L = \frac{5}{10^3} = 5\times10^{-3}\ \text{H} = 5\ \text{mH}\).

Teacher's Note:
1. Self-inductance equals the induced emf divided by the rate of change of current.
2. Convert milliseconds to seconds before dividing.

Answer: (d) \(\frac{\mu_0 N^2I^2b^2}{4R}\)

Explanation:
1. Field inside \(B = \frac{\mu_0 NI}{2\pi R}\); volume \(= \pi b^2\times 2\pi R\).
2. Energy \(U = \frac{B^2}{2\mu_0}\times\text{volume} = \frac{\mu_0 N^2I^2}{8\pi^2R^2}\times 2\pi^2 b^2 R\).
3. \(U = \frac{\mu_0 N^2I^2b^2}{4R}\). (Check: \(L = \frac{\mu_0N^2b^2}{2R}\) and \(U = \frac{1}{2}LI^2\).)

Teacher's Note:
1. Energy equals energy density times volume, or \(\frac{1}{2}LI^2\).
2. For \(R \gg b\) take the field as uniform over the cross-section.

Answer: (a) \(8.4\ \text{V}\)

Explanation:
1. Rotating through \(180^\circ\) reverses the flux: \(\Delta\phi = 2NBA = 2\times200\times0.3\times70\times10^{-4} = 0.84\ \text{Wb}\).
2. \(\varepsilon = \frac{\Delta\phi}{\Delta t} = \frac{0.84}{0.1} = 8.4\ \text{V}\).

Teacher's Note:
1. A half turn changes the flux from \(+NBA\) to \(-NBA\), a change of \(2NBA\).
2. Convert square centimetres to square metres.

Answer: (b) \(5\pi\)

Explanation:
1. \(\varepsilon = M\frac{dI}{dt} = MI_0\omega\cos\omega t\).
2. Maximum \(= 0.005\times10\times100\pi = 5\pi\ \text{V}\).

Teacher's Note:
1. The emf is the derivative of the sinusoidal current, so its amplitude is \(MI_0\omega\).
2. The maximum of the cosine factor is 1.

Answer: (a) \(450\ \text{A s}^{-1}\)

Explanation:
1. At \(t = 0\) the current is zero, so there is no voltage drop across the resistor and the entire emf appears across the inductor.
2. \(L\frac{dI}{dt} = V\), so \(\frac{dI}{dt} = \frac{90}{0.2} = 450\ \text{A/s}\).

Teacher's Note:
1. At the instant of switching on, an inductor behaves like an open circuit.
2. The resistor drop is zero because the current is still zero.

Answer: (a) \(\frac{e^2}{e^2 - 1}\)

Explanation:
1. Time constant \(\tau = \frac{L}{R} = 0.5\ \text{s}\), and \(I(t) = I_\infty\left(1 - e^{-t/\tau}\right)\).
2. \(I(1\ \text{s}) = I_\infty(1 - e^{-2})\).
3. \(\frac{I_\infty}{I(1)} = \frac{1}{1 - e^{-2}} = \frac{e^2}{e^2 - 1}\).

Teacher's Note:
1. Compute the time constant first, then use the growth law of current in an LR circuit.
2. Express the answer in the form given by the options.

Answer: (d) the pointer of the galvanometer will show a deflection

Explanation:
1. Switching on the current makes the flux through coil 2 change from zero to a non-zero value.
2. This induces an emf and a momentary current in coil 2, so the galvanometer pointer deflects.
3. The coils are fixed, so they cannot move, and the pointer does not remain at zero.

Teacher's Note:
1. Induced emf appears only while the flux is changing.
2. The pointer returns to zero once the current in coil 1 becomes steady.

Answer: (a) 0

Explanation:
1. Flux \(= BA\cos\theta\), where \(\theta\) is the angle between the field and the normal to the surface.
2. A field parallel to the surface is perpendicular to the normal, so \(\cos90^\circ = 0\).
3. The flux is zero.

Teacher's Note:
1. Flux counts the field lines crossing the surface.
2. Lines parallel to the surface do not cross it.

Answer: (d) \(0.05\ \text{A}\)

Explanation:
1. With no losses, input power \(=\) output power: \(V_pI_p = V_sI_s\).
2. \(I_s = \frac{220\times5}{22000} = 0.05\ \text{A}\).

Teacher's Note:
1. A step-up transformer raises the voltage and lowers the current.
2. Power is the same on both sides for an ideal transformer.

Answer: (c) \(16\ \text{J/s}\)

Explanation:
1. \(U = \frac{1}{2}LI^2\), so \(\frac{dU}{dt} = LI\frac{dI}{dt}\).
2. \(\frac{dU}{dt} = 2\times2\times4 = 16\ \text{J/s}\).

Teacher's Note:
1. The rate of increase of stored energy is the power absorbed by the inductor.
2. It equals the emf across the coil times the current.

Answer: (a) \(3\ \text{A}\)

Explanation:
1. Output power \(= 120\times20 = 2400\ \text{W}\).
2. Input power \(= \frac{2400}{0.8} = 3000\ \text{W}\).
3. Current drawn from the line \(= \frac{3000}{1000} = 3\ \text{A}\).

Teacher's Note:
1. Efficiency is output power divided by input power.
2. Find the input power first and then divide by the line voltage.

Answer: (d) \(20\)

Explanation:
1. Two inductors in parallel give \(\frac{L}{2}\) (no mutual coupling).
2. In series with \(5\ \text{mH}\): \(\frac{L}{2} + 5 = 15\).
3. \(L = 20\ \text{mH}\).

Teacher's Note:
1. Inductors combine like resistors when there is no mutual inductance.
2. Solve the resulting simple equation.

Answer: (a) \(25\ \text{H}\)

Explanation:
1. Mutual inductance \(M \propto N_1N_2\) for fixed geometry.
2. Before: \(5\times10 = 50\). After: \(10\times5 = 50\).
3. The product is unchanged, so \(M = 25\ \text{H}\).

Teacher's Note:
1. Mutual inductance depends on the product of the numbers of turns.
2. Swapping the primary and secondary leaves it unchanged.

Answer: (b) \(450\ \text{V},\ 15\ \text{A}\)

Explanation:
1. Primary current \(I_p = \frac{3000}{200} = 15\ \text{A}\).
2. Output power \(= 0.9\times3000 = 2700\ \text{W}\).
3. Secondary voltage \(V_s = \frac{2700}{6} = 450\ \text{V}\).

Teacher's Note:
1. Input power is the given supply power, and the output is efficiency times input.
2. The primary current uses the input power, not the output power.

Answer: (a) potential difference between P and Q is equal to \(2B\omega a^2\)

Explanation:
1. The emf depends only on the straight line joining P to Q, of length \(2a\), about which the wire rotates: \(\varepsilon = \frac{1}{2}B\omega(2a)^2 = 2B\omega a^2\).
2. For clockwise rotation with \(\vec B\) into the page, \(\vec v\times\vec B\) on the points of PQ points from P towards Q, so Q is at the higher potential and statement c is false.

Teacher's Note:
1. For a wire rotating about one end, use the chord joining its ends.
2. The shape of the wire between the ends does not change the emf.

Answer: (c) \(20\pi\ \text{mV}\)

Explanation:
1. \(\omega = 2\pi\times20 = 40\pi\ \text{rad/s}\).
2. \(\varepsilon = \frac{1}{2}B\omega r^2 = \frac{1}{2}\times0.1\times40\pi\times(0.1)^2\).
3. \(\varepsilon = 0.02\pi\ \text{V} = 20\pi\ \text{mV}\).

Teacher's Note:
1. Each element of the radius moves at speed \(\omega x\), so the average speed is \(\frac{\omega r}{2}\).
2. This gives the factor \(\frac{1}{2}\) in the formula.

Answer: (a) the current in each loop will decrease

Explanation:
1. Each loop produces flux through the other in the same sense as its own flux.
2. As the loops approach, the flux through each loop increases.
3. By Lenz's law each loop opposes the increase, so the current in each decreases.

Teacher's Note:
1. Lenz's law always opposes the change that causes the induction.
2. Currents in the same direction increase the linked flux as the loops come closer.

Answer: (a) \(12\ \text{mV}\)

Explanation:
1. \(\vec v\times\vec B = \hat j\times\hat k = \hat i\), so the electric field in the cube is along \(x\) and the potential difference is across the faces perpendicular to \(x\).
2. \(\varepsilon = Bvl = 0.1\times6\times0.02 = 0.012\ \text{V} = 12\ \text{mV}\).

Teacher's Note:
1. The separation of charge happens along \(\vec v\times\vec B\).
2. The relevant length is the edge of the cube in that direction.

Answer: (d) \(2\times10^{-3}\ \text{V}\)

Explanation:
1. \(\varepsilon = \frac{\Delta\phi}{\Delta t} = \frac{8\times10^{-4}}{0.4}\).
2. \(\varepsilon = 2\times10^{-3}\ \text{V}\).

Teacher's Note:
1. Faraday's law needs the change in flux and the time taken.
2. Check the power of ten in the answer.

Answer: (d) \(5\ \text{mH}\)

Explanation:
1. \(\varepsilon_{max} = M I_0\omega = M\times20\times100\pi\).
2. \(10\pi = 2000\pi M\), so \(M = 5\times10^{-3}\ \text{H} = 5\ \text{mH}\).

Teacher's Note:
1. The maximum emf is the mutual inductance times the amplitude of the rate of change of current.
2. The amplitude of the rate of change is \(I_0\omega\).

Answer: (c) if \(E\) is not zero, \(\phi\) may or may not be zero

Explanation:
1. \(E = -\frac{d\phi}{dt}\) depends on the rate of change of flux and not on its value.
2. The flux can be zero at an instant while changing quickly (so \(E \neq 0\)), and it can be non-zero but steady (so \(E = 0\)).
3. Hence if \(E \neq 0\), \(\phi\) may or may not be zero.

Teacher's Note:
1. Emf depends on how fast the flux changes, not on how large it is.
2. Think of a coil rotating in a field and the instant when it is perpendicular to the field lines.