JEE Mains · Physics · STD 12 - Moving charges and magnetism
- JEE Mains 2020
- Hard
- MCQ
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
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(-\frac{m v^{2} \vec{B}}{2 B^{2}}\)
Step-by-step Solution
Detailed explanation
Magnetic moment
\(M=i A\)
\(M=\left(\frac{q}{T}\right) \times \pi r^{2}=\frac{q \pi r^{2}}{\left(\frac{2 \pi r}{v}\right)}=\frac{q v r}{2}\)
\(M=\frac{q v}{2} \times \frac{v m}{q B}\)
\(M =\frac{ mv ^{2}}{2 B }\)
As we can see from the figure, direction of magnetic moment \((M)\) is opposite to magnetic field.
\(\overrightarrow{ M }=-\frac{ mv ^{2}}{2 B } \hat{ B }\)
\(=-\frac{m v^{2}}{2 B^{2}} \vec{B}\)
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