Work Done by Transverse Magnetic Field – Rankers Physics

Force Acting on Moving Charges: Practice Problem & Solution

A particle of mass \(M\) and charge \(Q\) moving with velocity \(\vec{v}\) describes a circular path of radius \(R\) when subjected to a uniform transverse magnetic field of induction \(B\). The work done by the field when the particle completes one full circle is:
\(BQv2\pi R\)
\(\left(\frac{M v^2}{R}\right) 2\pi R\)
Zero
\(BQ2\pi R\)

Solution Explained:

To solve this problem, we apply the core principles of Force Acting on Moving Charges. Understanding the underlying formula is key to arriving at the correct answer below:

The magnetic force \(\vec{F} = Q(\vec{v} \times \vec{B})\) is always perpendicular to the velocity \(\vec{v}\). Therefore, power \(P = \vec{F} \cdot \vec{v} = 0\), meaning the work done is always zero.

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