Magnetism and Matter — NEET UG practice

1. Diamagnetic materials are weakly repelled by a magnetic field, and their susceptibility is small and negative. This corresponds to the range.

$$0 > \chi \ge -1$$

So A matches with II.

2. Ferromagnetic materials get strongly magnetised and can retain magnetism, so their susceptibility is very large and positive.

$$\chi >> 1$$

So B matches with III.

3. Paramagnetic materials are weakly attracted by a magnetic field, giving a small positive susceptibility bounded by a small number $\varepsilon$.

$$0 < \chi < \varepsilon$$

So C matches with IV.

4. Non-magnetic materials neither attract nor repel field lines, so their susceptibility is exactly zero.

$$\chi = 0$$

So D matches with I.

Hence, the correct matching is A-II, B-III, C-IV, D-I, so the answer is Option A.

1. Statement A: A magnetic sheet placed near a strong magnetic pole experiences an attractive (or repulsive) magnetic force. To keep it stationary against this pull, an external force is required. So statement A is true.

2. Statement B: A non-magnetic sheet does not interact with the magnetic field at all, so no force acts on it and none is needed to hold it in place. So statement B is false.

3. Statement C: Moving a conducting sheet away from the pole changes the magnetic flux through it. By Faraday's law of electromagnetic induction, this changing flux induces eddy currents, and by Lenz's law, these currents oppose the motion. So an external force is needed to move it at uniform velocity, making statement C true.

4. Statement D: A sheet that is both non-conducting and non-magnetic (non-polar) has no eddy currents and no magnetic interaction with the pole, so no opposing force exists, and hence no extra force is needed to move it at uniform velocity. So statement D is false.

Since only A and C hold true, the answer is Option B, A and C only.

1. Before bending, the magnetic moment of the straight bar of length $L$ and pole strength $m$ is given by.

$$M = mL$$

2. When bent at the middle so that the two halves (each of length $L/2$) make an angle of $60^\circ$ with each other, the new magnetic moment depends on the straight-line distance between the two end poles, not the bent path. This effective distance can be found using the two arms and the angle between them.

$$\Delta l = 2\left(\frac{L}{2}\right)\sin 30^\circ$$

3. Since $\sin 30^\circ = \frac{1}{2}$, simplifying the above gives the new effective length.

$$\Delta l = \frac{L}{2}$$

4. So the new magnetic moment is pole strength times this new effective length, and substituting $mL = M$ gives the final relation.

$$M' = m\cdot\frac{L}{2} = \frac{M}{2}$$

Hence, the answer is Option B, $\frac{M}{2}$.