JEE MAIN - Physics (2020 - 6th September Evening Slot - No. 9)
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:
$${{m{v^2}\overrightarrow B } \over {2{B^2}}}$$
-$${{m{v^2}\overrightarrow B } \over {2{B^2}}}$$
-$${{m{v^2}\overrightarrow B } \over {{B^2}}}$$
-$${{m{v^2}\overrightarrow B } \over {2\pi {B^2}}}$$
Explanation
_6th_September_Evening_Slot_en_9_1.png)
Magnetic dipole moment
M = iA
$$ \Rightarrow $$ M = $$\left( {{q \over T}} \right) \times \pi {R^2}$$
= $${{{q\pi {R^2}} \over {\left( {{{2\pi r} \over v}} \right)}}}$$ = $${{qvR} \over 2}$$
$$ \Rightarrow $$ M = $${{qv} \over 2} \times {{vm} \over {qB}}$$
As you can see from the figure, direction of magnetic moment(M) is opposite to magnetic field.
$$ \therefore $$ $$\overrightarrow M = - {{m{v^2}} \over {2B}}\widehat B$$
= $$ - {{m{v^2}} \over {2B}}\left( {{{\overrightarrow B } \over B}} \right)$$
= $$ - {{m{v^2}} \over {2{B^2}}}\overrightarrow B $$
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