JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism

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

A \(-\frac{ mv ^{2} \overrightarrow{ B }}{ B ^{2}}\)
B \(-\frac{m v^{2} \vec{B}}{2 \pi B^{2}}\)
C \(\frac{m v^{2} \vec{B}}{2 B^{2}}\)
D \(-\frac{m v^{2} \vec{B}}{2 B^{2}}\)

Verified Solution

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…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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