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JEE Advance - Physics (2013 - Paper 2 Offline - No. 15)

A point charge Q is moving in a circular orbit of radius R in the xy-plane with an angular velocity $$\omega$$. This can be considered as equivalent to a loop carrying a steady current $${{Q\omega } \over {2\pi }}$$. A uniform magnetic field along the positive z-axis is now switched on, which increases at a constant rate from 0 to B in one second. Assume that the radius of the orbit remains constant. The application of the magnetic field induces an emf in the orbit. The induced emf is defined as the work done by an induced electric field in moving a unit positive charge around a closed loop. It is known that, for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum with a proportionality constant $$\gamma$$.
A point charge Q is moving in a circular orbit of radius R in the xy-plane with an angular velocity $$\omega$$. This can be considered as equivalent to a loop carrying a steady current $${{Q\omega } \over {2\pi }}$$. A uniform magnetic field along the positive z-axis is now switched on, which increases at a constant rate from 0 to B in one second. Assume that the radius of the orbit remains constant. The application of the magnetic field induces an emf in the orbit. The induced emf is defined as the work done by an induced electric field in moving a unit positive charge around a closed loop. It is known that, for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum with a proportionality constant $$\gamma$$.
A point charge Q is moving in a circular orbit of radius R in the xy-plane with an angular velocity $$\omega$$. This can be considered as equivalent to a loop carrying a steady current $${{Q\omega } \over {2\pi }}$$. A uniform magnetic field along the positive z-axis is now switched on, which increases at a constant rate from 0 to B in one second. Assume that the radius of the orbit remains constant. The application of the magnetic field induces an emf in the orbit. The induced emf is defined as the work done by an induced electric field in moving a unit positive charge around a closed loop. It is known that, for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum with a proportionality constant $$\gamma$$.
The change in the magnetic dipole moment associated with the orbit, at the end of the time interval of the magnetic field change, is
$$ - \gamma BQ{R^2}$$
$$ - \gamma {{BQ{R^2}} \over 2}$$
$$\gamma {{BQ{R^2}} \over 2}$$
$$\gamma BQ{R^2}$$

Explanation

$${M \over L} = {Q \over {2m}}$$

JEE Advanced 2013 Paper 2 Offline Physics - Magnetism Question 24 English Explanation

$$\therefore$$ $$M = \left( {{Q \over {2m}}} \right)L$$

$$ \Rightarrow M \propto L$$, where $$\gamma = {Q \over {2m}}$$

$$\left( {{Q \over {2m}}} \right)(I\omega )$$

$$ = \left( {{Q \over {2m}}} \right)(m{R^2}\omega ) = {{Q\omega {R^2}} \over 2}$$

Induced electric field is opposite. Therefore,

$$\omega ' = \omega - \alpha t$$

$$\alpha = {\tau \over I} = {{(QE)R} \over {m{R^2}}} = {{(Q)\left( {{{BR} \over 2}} \right)R} \over {m{R^2}}} = {{QB} \over {2m}}$$

$$\therefore$$ $$\omega ' = \omega - {{QB} \over {2m}}.1 = \omega - {{QB} \over {2m}}$$

$${M_f} = {{Q\omega '{R^2}} \over 2} = Q\left( {\omega - {{QB} \over {2m}}} \right){{{R^2}} \over 2}$$

$$\therefore$$ $$\Delta M = {M_f} - {M_i} = - {{{Q^2}B{R^2}} \over {4m}}$$

$$M = - \gamma {{QB{R^2}} \over 2}$$ (as $$\gamma = {Q \over {2m}}$$)

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