JEE Physics Moving Charges and Magnetism MCQs Set 25

Answer: (b) \(3\times10^{-4}\ \text{N}\) towards right

Explanation:
1. The middle current is opposite to both outer currents, so both outer wires repel it.
2. Force from \(I_1\) (away from \(I_1\), to the right): \(F_1 = \frac{\mu_0 I_1 I L}{2\pi d_1} = \frac{2\times10^{-7}\times30\times10\times0.25}{0.03} = 5\times10^{-4}\ \text{N}\).
3. Force from \(I_2\) (away from \(I_2\), to the left): \(F_2 = \frac{2\times10^{-7}\times20\times10\times0.25}{0.05} = 2\times10^{-4}\ \text{N}\).
4. Net force \(= 5\times10^{-4} - 2\times10^{-4} = 3\times10^{-4}\ \text{N}\) to the right.

Teacher's Note:
1. Parallel currents attract and antiparallel currents repel.
2. Add the forces vectorially after deciding the direction of each.

Answer: (b) towards the wire

Explanation:
1. Take the current along \(+z\); at a point on the \(+x\) side the field is along \(+y\).
2. Force on the negative charge \(\vec F = -e(\vec v\times\vec B)\) must be along \(+z\), so \(\vec v\times\hat y\) must be along \(-z\).
3. This needs \(\vec v\) along \(-x\), that is towards the wire.

Teacher's Note:
1. For a negative charge the force is opposite to \(\vec v\times\vec B\).
2. Choose axes and check the direction of the cross product.

Answer: (c) (i) soft iron and (ii) steel respectively

Explanation:
1. The width of the loop along the H-axis measures coercivity.
2. Soft iron has low coercivity, so it has a narrow loop.
3. Steel has high coercivity and retentivity, so it has a wide loop.
4. Diamagnetic and paramagnetic substances show no hysteresis loop at all.

Teacher's Note:
1. Narrow loop means easy to magnetise and demagnetise (soft iron, used in electromagnets and transformers).
2. Wide loop means permanent magnet material such as steel.

Answer: (d) \(\sqrt{3}\)

Explanation:
1. For a tangent galvanometer \(\frac{\mu_0 n I}{2R} = B_H\tan\theta\), so \(n \propto \tan\theta\) for the same \(I\) and \(R\).
2. \(\frac{n_1}{n_2} = \frac{\tan 60^\circ}{\tan 45^\circ} = \frac{\sqrt{3}}{1} = \sqrt{3}\).

Teacher's Note:
1. In series the current through both coils is the same.
2. Compare the two instruments using the tangent law.

Answer: (c) \(\left(\frac{\mu_0 i}{2\pi R^2}\right) r\)

Explanation:
1. For a uniform current density, the current enclosed within radius \(r\) is \(i' = i\frac{r^2}{R^2}\).
2. Ampere's law: \(B(2\pi r) = \mu_0 i\frac{r^2}{R^2}\).
3. \(B = \frac{\mu_0 i r}{2\pi R^2}\).

Teacher's Note:
1. Only the current inside the Amperian loop contributes.
2. The field inside a solid wire grows linearly with the distance from the axis.

Answer: (c) \(\varepsilon_r = 1.5,\ \mu_r = 0.5\)

Explanation:
1. A diamagnetic material has \(\mu_r\) slightly less than 1, so \(\mu_r = 0.5\) is the only acceptable value among the choices that is below 1.
2. Relative permittivity of any material is greater than 1, so \(\varepsilon_r = 1.5\) is acceptable and \(0.5\) is not.

Teacher's Note:
1. Relative permittivity is always at least 1.
2. Diamagnetic means relative permeability below 1.

Answer: (d) \(\frac{x}{8}\)

Explanation:
1. \(B = \frac{\mu_0 I}{2R}\) and \(M = I\pi R^2\), so \(\frac{B}{M} = \frac{\mu_0}{2\pi R^3}\).
2. The current cancels, and the ratio varies as \(\frac{1}{R^3}\).
3. Doubling \(R\) gives \(\frac{x}{8}\).

Teacher's Note:
1. Simplify the ratio symbolically before putting in the changes.
2. Any quantity that cancels can be ignored.

Answer: (d) \(5\ \text{A m}^2\)

Explanation:
1. The magnets are perpendicular to each other, so the moments add as vectors.
2. \(M = \sqrt{4^2 + 3^2} = 5\ \text{A m}^2\).

Teacher's Note:
1. A cross arrangement means the angle between the moments is \(90^\circ\).
2. Use the Pythagoras result for perpendicular vectors.

Answer: (a) \(\theta = 0^\circ\)

Explanation:
1. \(\tau = MB\sin\theta\), so \(\frac{d\tau}{d\theta} = MB\cos\theta\).
2. This is maximum when \(\cos\theta = 1\), that is \(\theta = 0^\circ\).

Teacher's Note:
1. Rate of change of a function is its derivative.
2. Cosine has its largest value at zero angle.

Answer: (b) \(\frac{\mu_0 i}{4R}\left(\frac{2}{\pi}+1\right)\)

Explanation:
1. Semicircle: \(B_1 = \frac{\mu_0 i}{4R}\).
2. Each semi-infinite straight wire, with P at the perpendicular from its end, gives \(B = \frac{\mu_0 i}{4\pi R}\).
3. All three contributions point the same way (out of the page for the current direction shown).
4. Total \(B = \frac{\mu_0 i}{4R} + 2\cdot\frac{\mu_0 i}{4\pi R} = \frac{\mu_0 i}{4R}\left(1 + \frac{2}{\pi}\right)\).

Teacher's Note:
1. A semi-infinite wire seen from the foot of the perpendicular gives half the field of an infinite wire.
2. Check the directions with the right-hand rule before adding.

Answer: (a) in the direction of the magnetic dipole moment \(\vec M\) of the magnet

Explanation:
1. On the axial line, \(\vec B = \frac{\mu_0}{4\pi}\frac{2\vec M}{z^3}\).
2. The field is therefore along \(\vec M\), that is from the south pole to the north pole inside the magnet and continuing in the same sense outside on the axis.

Teacher's Note:
1. The axial field is parallel to the dipole moment, and the equatorial field is antiparallel to it.
2. The magnitude of M changes the strength, not the direction.

Answer: (b) \(0.0120\)

Explanation:
1. Curie's law: \(\chi \propto \frac{1}{T}\).
2. \(T_1 = 200\ \text{K}\) and \(T_2 = 100\ \text{K}\).
3. \(\chi_2 = 0.0060\times\frac{200}{100} = 0.0120\).

Teacher's Note:
1. Always convert Celsius to kelvin in Curie's law.
2. Halving the temperature doubles the susceptibility.

Answer: (d) a proton and an \(\alpha\) particle

Explanation:
1. The radius is \(r = \frac{mv}{qB}\), so for the same velocity and field \(r \propto \frac{m}{q}\).
2. Both particles are positive, which rules out the electron.
3. \(M\) has the smaller \(\frac{m}{q}\): proton \(\frac{m}{e}\); deuteron \(\frac{2m}{e}\); \(\alpha\) particle \(\frac{4m}{2e} = \frac{2m}{e}\).
4. A proton (smaller radius) and a heavier-per-charge particle such as an \(\alpha\) particle (larger radius) fit; a deuteron then proton would reverse the order.

Teacher's Note:
1. For equal speeds the radius is proportional to the mass-to-charge ratio.
2. The direction of curvature tells you the sign of the charge.

Answer: (b) \(5\times10^{-4}\)

Explanation:
1. \(B = \frac{\mu_0 N i}{2\pi r} = \frac{2\times10^{-7}\times500\times0.5}{0.1}\).
2. \(B = 5\times10^{-4}\ \text{T}\).

Teacher's Note:
1. A toroid has a field like a solenoid with \(n = \frac{N}{2\pi r}\).
2. The field is almost uniform along the circular axis.

Answer: (a) \(1\ \text{s}\)

Explanation:
1. \(T = 2\pi\sqrt{\frac{I}{MB}}\), and the magnetic moment \(M = m\times(2l)\) is proportional to the pole strength.
2. Pole strength four times means \(M\) is four times, so \(T\) becomes \(\frac{1}{\sqrt{4}} = \frac{1}{2}\) of the original.
3. New period \(= 1\ \text{s}\).

Teacher's Note:
1. Period varies as the inverse square root of the magnetic moment.
2. The moment of inertia stays unchanged.

Answer: (a) \(5\times10^{-4}\ \text{N}\) (attraction)

Explanation:
1. \(F = \frac{\mu_0 I_1 I_2 L}{2\pi d} = \frac{2\times10^{-7}\times10\times15\times5}{0.3}\).
2. \(F = 5\times10^{-4}\ \text{N}\).
3. The currents are in the same direction, so the force is attractive.

Teacher's Note:
1. Same direction means attraction; opposite directions mean repulsion.
2. Convert the distance to metres.

Answer: (d) zero

Explanation:
1. The current divides equally between the two semicircles, \(\frac{i}{2}\) in each.
2. The two semicircles produce magnetic fields of equal magnitude and opposite direction at the centre, so the net field there is zero.
3. With \(B = 0\) the magnetic force is zero.

Teacher's Note:
1. Two identical halves carrying the same current in opposite senses cancel at the centre.
2. No field means no magnetic force whatever the speed.

Answer: (a) material B

Explanation:
1. Susceptibility \(\chi = \frac{I}{H}\) is the slope of the \(I\)-\(H\) line.
2. Line B is steeper than line A.
3. So material B has the larger susceptibility.

Teacher's Note:
1. Susceptibility is the ratio of magnetisation to magnetising field.
2. On a straight-line graph it is the slope.

Answer: (a) a force and a torque

Explanation:
1. The two poles are in different field strengths, so the forces on them are unequal and the net force is not zero.
2. The poles are generally not aligned with the field, so there is also a torque.

Teacher's Note:
1. In a uniform field a needle feels only a torque.
2. In a non-uniform field there is a force as well.

Answer: (b) \(\frac{\epsilon_0 l B}{\sigma}\)

Explanation:
1. The field between the plates is \(E = \frac{\sigma}{\epsilon_0}\).
2. Constant velocity means the electric and magnetic forces balance: \(eE = evB\), so \(v = \frac{E}{B} = \frac{\sigma}{\epsilon_0 B}\).
3. Time to cross \(t = \frac{l}{v} = \frac{\epsilon_0 l B}{\sigma}\).

Teacher's Note:
1. Straight-line motion in crossed fields means the velocity equals the ratio of the fields.
2. The charge of the electron cancels from the balance.

Answer: (c) \(\frac{\mu_0 q\omega}{4\pi R}\)

Explanation:
1. The rotating charge is equivalent to a current \(I = \frac{q}{T} = \frac{q\omega}{2\pi}\).
2. Field at the centre of a circular current \(B = \frac{\mu_0 I}{2R}\).
3. \(B = \frac{\mu_0}{2R}\cdot\frac{q\omega}{2\pi} = \frac{\mu_0 q\omega}{4\pi R}\).

Teacher's Note:
1. A rotating charged ring is a current loop.
2. Current equals charge divided by the time of one revolution.

Answer: (d) a ferromagnetic substance becomes paramagnetic

Explanation:
1. At the Curie temperature the domain structure of a ferromagnet breaks down due to thermal agitation.
2. Above it the substance behaves as a paramagnetic substance and follows the Curie-Weiss law.

Teacher's Note:
1. Ferromagnetism disappears when heat destroys the domains.
2. The other changes listed do not occur.

Answer: (d) \(q\vec E + q(\vec v\times\vec B)\)

Explanation:
1. The electric force is \(q\vec E\) and the magnetic force is \(q(\vec v\times\vec B)\).
2. The total force is their sum, the Lorentz force \(\vec F = q\vec E + q(\vec v\times\vec B)\).

Teacher's Note:
1. The order in the cross product matters, since reversing it changes the sign.
2. The electric force is independent of velocity.

Answer: (d) a steady attractive force

Explanation:
1. The observer looks at the two coils from opposite directions, because he is between them.
2. A current which is counterclockwise when seen from one side is clockwise when seen from the other side.
3. So both currents circulate in the same sense, and parallel currents attract each other, giving a steady attractive force.

Teacher's Note:
1. A coil's sense of rotation depends on the side from which you look at it.
2. Coils with currents in the same sense attract, and with opposite senses repel.