NEET Magnetic Effects of Current MCQs & PYQs

Q: 18. The work done in turning a magnet of magnetic moment $M$ by an angle of $90^{\circ}$ from the meridian, is $n$ times the corresponding work done to turn it through an angle of $60^{\circ}$. The value of $n$ is given by (1995)

$W_{90^{\circ}} = MB(1 - \cos 90^{\circ}) = MB(1 - 0) = MB$. $W_{60^{\circ}} = MB(1 - \cos 60^{\circ}) = MB(1 - 0.5) = 0.5MB$. Since $W_{90^{\circ}} = n W_{60^{\circ}}$, we have $MB = n(0.5MB) \Rightarrow n = 2$.

Q: 19. An iron rod of susceptibility $599$ is subjected to a magnetising field of $1200$ A m$^{-1}$. The permeability of the material of the rod is : (2020) ($\mu_0 = 4\pi \times 10^{-7}$ T m A$^{-1}$)

Relative permeability $\mu_r = 1 + \chi = 1 + 599 = 600$. Permeability $\mu = \mu_r \mu_0 = 600 \times 4\pi \times 10^{-7} = 2400\pi \times 10^{-7} = 2.4\pi \times 10^{-4}$ T m A$^{-1}$.

Q: 7. A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in 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: (2016 - II)

Work done, $W = MB(1 - \cos 60^{\circ}) = \frac{MB}{2} \Rightarrow MB = 2W$. Torque required, $\tau = MB \sin 60^{\circ} = 2W \times \frac{\sqrt{3}}{2} = \sqrt{3}W$.

Q: 8. A magnetic needle suspended parallel to a magnetic field requires $\sqrt{3}$ J of work to turn it through $60^{\circ}$. The torque needed to maintain the needle in this position will be: (2012 Mains)

$W = MB(1 - \cos 60^{\circ}) = \frac{MB}{2} = \sqrt{3} \Rightarrow MB = 2\sqrt{3}$ J. Torque, $\tau = MB \sin 60^{\circ} = 2\sqrt{3} \times \frac{\sqrt{3}}{2} = 3$ J.

Q: 9. A short bar magnet of magnetic moment $0.4$ J T$^{-1}$ is placed in a uniform magnetic field of $0.16$ T. The magnet is in stable equilibrium when the potential energy is: (2011 Mains)

For stable equilibrium, the angle between $M$ and $B$ is $0^{\circ}$. Potential energy $U = -MB \cos 0^{\circ} = - (0.4)(0.16)(1) = -0.064$ J.

Q: 12. A compass needle which is allowed to move in a horizontal plane is taken to a geomagnetic pole. It: (2012 Pre)

At the geomagnetic poles, the Earth's magnetic field is entirely vertical, meaning the horizontal component $B_H$ is zero. A compass free to rotate only horizontally experiences no directing torque and will stay in any position.

Practice NEET Magnetic Effects of Current Questions

Question 241:

easy

14. A vibration magnetometer placed in magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of $2$ sec in earth’s horizontal magnetic field of $24$ microtesla. When a horizontal field of $18$ microtesla is produced opposite to the earth’s field by placing a current carrying wire, the new time period of magnet will be: (2010 Pre)

Initial field $B_1 = 24$ $\mu$T, $T_1 = 2$ s. Net new field $B_2 = 24 - 18 = 6$ $\mu$T.
Since $T \propto \frac{1}{\sqrt{B}}$, we have $\frac{T_2}{T_1} = \sqrt{\frac{B_1}{B_2}} = \sqrt{\frac{24}{6}} = \sqrt{4} = 2$. Therefore, $T_2 = 2 \times 2 = 4$ s.

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Question 242:

easy

15. A bar magnet having a magnetic moment of $2 \times 10^4$ J T$^{-1}$ is free to rotate in a horizontal plane. A horizontal magnetic field $B = 6 \times 10^{-4}$ T exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction $60^{\circ}$ from the field is: (2009)

Work done, $W = MB(\cos \theta_1 - \cos \theta_2) = MB(\cos 0^{\circ} - \cos 60^{\circ}) = MB(1 - \frac{1}{2}) = \frac{MB}{2}$.
$W = \frac{1}{2} \times (2 \times 10^4) \times (6 \times 10^{-4}) = 6$ J.

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Question 243:

easy

16. A bar magnet is oscillating in the Earth’s magnetic field with a period $T$. What happens to its period and motion if its mass is quadrupled? (2003)

Time period is given by $T = 2\pi \sqrt{\frac{I}{MB}}$. Moment of inertia $I$ is directly proportional to mass $m$. If mass is quadrupled, $I$ becomes $4I$.
New period $T' = 2\pi \sqrt{\frac{4I}{MB}} = 2T$. The restoring torque is still $-MB \sin \theta$, so motion remains S.H.M.

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Question 244:

easy

17. Two bar magnets having same geometry with magnetic moments $M$ and $2M$, are firstly placed in such a way that their similar poles are same side then its time period of oscillation is $T_1$. Now the polarity of one of the magnet is reversed then time period of oscillation is $T_2$, then: (2002)

$T = 2\pi \sqrt{\frac{I}{M_{net} B_H}}$. Initially $M_{net} = 2M + M = 3M$, so $T_1 \propto \frac{1}{\sqrt{3M}}$.
After reversing polarity, $M_{net} = 2M - M = M$, so $T_2 \propto \frac{1}{\sqrt{M}}$. Therefore, $T_1 < T_2$.

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Question 245:

easy

18. The work done in turning a magnet of magnetic moment $M$ by an angle of $90^{\circ}$ from the meridian, is $n$ times the corresponding work done to turn it through an angle of $60^{\circ}$. The value of $n$ is given by (1995)

$W_{90^{\circ}} = MB(1 - \cos 90^{\circ}) = MB(1 - 0) = MB$.
$W_{60^{\circ}} = MB(1 - \cos 60^{\circ}) = MB(1 - 0.5) = 0.5MB$.
Since $W_{90^{\circ}} = n W_{60^{\circ}}$, we have $MB = n(0.5MB) \Rightarrow n = 2$.

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Question 246:

easy

19. An iron rod of susceptibility $599$ is subjected to a magnetising field of $1200$ A m$^{-1}$. The permeability of the material of the rod is : (2020)
($\mu_0 = 4\pi \times 10^{-7}$ T m A$^{-1}$)

Relative permeability $\mu_r = 1 + \chi = 1 + 599 = 600$.
Permeability $\mu = \mu_r \mu_0 = 600 \times 4\pi \times 10^{-7} = 2400\pi \times 10^{-7} = 2.4\pi \times 10^{-4}$ T m A$^{-1}$.

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Question 247:

easy

7. A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in 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: (2016 – II)

Work done, $W = MB(1 - \cos 60^{\circ}) = \frac{MB}{2} \Rightarrow MB = 2W$.
Torque required, $\tau = MB \sin 60^{\circ} = 2W \times \frac{\sqrt{3}}{2} = \sqrt{3}W$.

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Question 248:

easy

8. A magnetic needle suspended parallel to a magnetic field requires $\sqrt{3}$ J of work to turn it through $60^{\circ}$. The torque needed to maintain the needle in this position will be: (2012 Mains)

$W = MB(1 - \cos 60^{\circ}) = \frac{MB}{2} = \sqrt{3} \Rightarrow MB = 2\sqrt{3}$ J.
Torque, $\tau = MB \sin 60^{\circ} = 2\sqrt{3} \times \frac{\sqrt{3}}{2} = 3$ J.

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Question 249:

easy

9. A short bar magnet of magnetic moment $0.4$ J T$^{-1}$ is placed in a uniform magnetic field of $0.16$ T. The magnet is in stable equilibrium when the potential energy is: (2011 Mains)

For stable equilibrium, the angle between $M$ and $B$ is $0^{\circ}$.
Potential energy $U = -MB \cos 0^{\circ} = - (0.4)(0.16)(1) = -0.064$ J.

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Question 250:

easy

12. A compass needle which is allowed to move in a horizontal plane is taken to a geomagnetic pole. It: (2012 Pre)

At the geomagnetic poles, the Earth's magnetic field is entirely vertical, meaning the horizontal component $B_H$ is zero. A compass free to rotate only horizontally experiences no directing torque and will stay in any position.

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