Q. No.1. \(0.064~\text{J}\)
2. \(-0.064~\text{J}\)
3. zero
4. \(-0.082~\text{J}\)
A bar magnet of magnetic moment \(1.5~\text{J/T}\) lies aligned with the direction of a uniform magnetic field of \(0.22~\text{T}\). What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment normal to the field direction?
1. \(0.66\) J
2. \(0.33\) J
3. \(0\)
4. \(0.44\) J
A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in the equilibrium state. The energy required to rotate it by \(60^{\circ}\) is \(W\). Now the torque required to keep the magnet in this new position is:
| 1. | \(\dfrac{W}{\sqrt{3}}\) | 2. | \(\sqrt{3}W\) |
| 3. | \(\dfrac{\sqrt{3}W}{2}\) | 4. | \(\dfrac{2W}{\sqrt{3}}\) |
| Assertion (A): | Magnetic dipoles, when placed near a long straight current-carrying wire, are deflected perpendicular to the wire. |
| Reason (R): | The magnetic field lines of a straight current-carrying wire are oriented in a direction which is perpendicular to the wire. |
| 1. | Both (A) and (R) are True and (R) is the correct explanation of (A). |
| 2. | Both (A) and (R) are True but (R) is not the correct explanation of (A). |
| 3. | (A) is True but (R) is False. |
| 4. | (A) is False but (R) is True. |
The work done in rotating a magnet of the magnetic moment \(100\) A-m2 through \(90^\circ\) from a direction parallel to the uniform magnetic field of strength \(0.4 \times 10^{-4}\) Tesla is:
1. \(4\) mJ
2. zero
3. \(6\) mJ
4. \(8\) mJ
The magnetic field at the centre of a circular loop of area \(A\) is \(B.\) The magnetic moment of the loop is:
| 1. | \(\dfrac{BA^2}{\mu_0\pi}\) | 2. | \(\dfrac{BA\sqrt A}{\mu_0}\) |
| 3. | \(\dfrac{BA\sqrt A}{\mu_0\pi}\) | 4. | \(\dfrac{2BA\sqrt A}{\mu_0\sqrt\pi}\) |
| 1. | \(mB\) | 2. | \(2mB\) |
| 3. | \(\dfrac12mB\) | 4. | zero |
The ratio of the magnitudes of the equatorial and axial fields due to a bar magnet of length \(5.0~\text{cm}\) at a distance of \(50~\text{cm}\) from its mid-point is:
(given, the magnetic moment of the bar magnet is \(0.40~\text{Am}^{2}\))
1. \(\dfrac{1}{2}\)
2. \(2\)
3. \(1\)
4. \(\dfrac{3}{2}\)
A magnetic needle suspended parallel to a magnetic field requires \(\sqrt{3}~\text{J}\) of work to turn it through \(60^\circ\)
1. \(3\) N-m
2. \(\sqrt{3} \) N-m
3. \(\frac32\) N-m
4. \(2\sqrt{3}\) N-m
\(B_1=\dfrac{\mu_0}{4\pi}\dfrac{m}{r^3}(3\cos^2\theta-1)\\ B_2=\dfrac{\mu_0}{4\pi}\dfrac{m}{r^3}(3\sin\theta\cos\theta)\)
where, \(r=OP=\) the distance of \(P\) from the magnet's centre & \(\theta\) is the angle made by \(OP\) with the axis.
Three short identical bar magnets (moment \(m,\) each) are placed at the three vertices \(A,B,C\) of an equilateral triangle \(\triangle ABC,\) with their axes parallel to the base \(BC.\) The magnetic field at the point \(O~(OB=r)\) is:
| 1. | \(\left(\dfrac{\mu_0}{4\pi}\dfrac{m}{r^3}\right)\cdot\dfrac32 \) | 2. | \(\left(\dfrac{\mu_0}{4\pi}\dfrac{m}{r^3}\right)\cdot\dfrac12 \) |
| 3. | \(\left(\dfrac{\mu_0}{4\pi}\dfrac{m}{r^3}\right)\cdot\dfrac34 \) | 4. | \(\left(\dfrac{\mu_0}{4\pi}\dfrac{m}{r^3}\right)\cdot\dfrac14 \) |
© 2026 GoodEd Technologies Pvt. Ltd.