Question 171: Magnetism and Matter

Question

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:

Accepted Answer

Magnetic dipole moment
M = iA
$$ \Rightarrow $$ M = $$\left( {{q \over T}} ight) 	imes \pi {R^2}$$
= $${{{q\pi {R^2}} \over {\left( {{{2\pi r} \over v}} ight)}}}$$ = $${{qvR} \over 2}$$
$$ \Rightarrow $$ M = $${{qv} \over 2} 	imes {{vm} \over {qB}}$$
As you can see from the figure, direction of magnetic moment(M) is opposite to magnetic field.
$$ 	herefore $$ $$\overrightarrow M  =  - {{m{v^2}} \over {2B}}\widehat B$$
= $$ - {{m{v^2}} \over {2B}}\left( {{{\overrightarrow B } \over B}} ight)$$
= $$ - {{m{v^2}} \over {2{B^2}}}\overrightarrow B $$

Answer

$${{m{v^2}\overrightarrow B } \over {2{B^2}}}$$

Answer

-$${{m{v^2}\overrightarrow B } \over {2{B^2}}}$$

Answer

-$${{m{v^2}\overrightarrow B } \over {{B^2}}}$$

Answer

-$${{m{v^2}\overrightarrow B } \over {2\pi {B^2}}}$$

Magnetism and Matter

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:

Choose the correct answer: