A charged particle going around in a circle can be considered to be a current loop. A particle of mass m carrying charge q is moving in a plane with speed v under the influence of magnetic field \overrightarrow B . The magnetic moment of this moving particle:

- mv^2\overrightarrow B \over 2B^2

mv^2\overrightarrow B \over 2B^2

- mv^2\overrightarrow B \over B^2

- mv^2\overrightarrow B \over 2\pi B^2

Magnetics PYQ — JEE Main 2020 · 6 Sep · Shift 2 · Q44 with Solution

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Correct answer: B

Motion of the charged particle in a magnetic field

A charged particle moving perpendicular to a uniform magnetic field executes circular motion.

The magnetic force provides the centripetal force:

qvB = \frac{mv^2}{r}

So,

r = \frac{mv}{qB}

Current due to revolving charge

A charge completing one revolution in time period $T$ constitutes a current

I = \frac{q}{T}

Now,

T = \frac{2\pi r}{v}

Hence,

I = \frac{qv}{2\pi r}

Magnetic moment of the current loop

Magnetic moment magnitude is

\mu = IA

where area of the circular loop is

A = \pi r^2

Therefore,

\mu = \frac{qv}{2\pi r}\cdot \pi r^2 = \frac{qvr}{2}

Substitute $r = \frac{mv}{qB}$ :

\mu = \frac{qv}{2}\cdot \frac{mv}{qB} = \frac{mv^2}{2B}

So the magnitude is

\mu = \frac{mv^2}{2B}

Direction of magnetic moment

For a positive charge, $\vec\mu$ is along the angular momentum direction. But here the motion is caused by magnetic field, and using the relation between orbital magnetic moment and angular momentum:

\vec\mu = \frac{q}{2m}\vec L

Also, in circular motion under magnetic field, the angular momentum vector turns out to be opposite to $\vec B$ for positive charge, and because the magnetic moment associated with the orbit is opposite to $\vec B$ overall for the circulating charged particle in this setup.

Thus,

\vec\mu = -\frac{mv^2}{2B}\,\hat B

Writing in vector form,

\vec\mu = -\frac{mv^2\,\vec B}{2B^2}

Match with options

This corresponds to:

-\frac{m v^2 \vec B}{2B^2}

which is Option B.

Final Answer

Option B:

\boxed{\vec\mu = -\frac{m v^2 \vec B}{2B^2}}

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View written solutionFree

Correct answer: B

Motion of the charged particle in a magnetic field

A charged particle moving perpendicular to a uniform magnetic field executes circular motion.

The magnetic force provides the centripetal force:

qvB = \frac{mv^2}{r}

So,

r = \frac{mv}{qB}

Current due to revolving charge

A charge completing one revolution in time period $T$ constitutes a current

I = \frac{q}{T}

Now,

T = \frac{2\pi r}{v}

Hence,

I = \frac{qv}{2\pi r}

Magnetic moment of the current loop

Magnetic moment magnitude is

\mu = IA

where area of the circular loop is

A = \pi r^2

Therefore,

\mu = \frac{qv}{2\pi r}\cdot \pi r^2 = \frac{qvr}{2}

Substitute $r = \frac{mv}{qB}$ :

\mu = \frac{qv}{2}\cdot \frac{mv}{qB} = \frac{mv^2}{2B}

So the magnitude is

\mu = \frac{mv^2}{2B}

Direction of magnetic moment

For a positive charge, $\vec\mu$ is along the angular momentum direction. But here the motion is caused by magnetic field, and using the relation between orbital magnetic moment and angular momentum:

\vec\mu = \frac{q}{2m}\vec L

Thus,

\vec\mu = -\frac{mv^2}{2B}\,\hat B

Writing in vector form,

\vec\mu = -\frac{mv^2\,\vec B}{2B^2}

Match with options

This corresponds to:

-\frac{m v^2 \vec B}{2B^2}

which is Option B.

Final Answer

Option B:

\boxed{\vec\mu = -\frac{m v^2 \vec B}{2B^2}}

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