MHT CET · Physics · Magnetic Effects of Current

An electron moves in a circular orbit with uniform speed ' \(v^{\prime}\). It produces a
magnetic field 'B' at the centre of the circle. The radius of the circle is
\(\left[\mu_{0}=\right.\) permeability of free space, \(\mathrm{e}=\) electronic charge \(]\)

A \(\left(\frac{\mu_{0} \mathrm{ev}}{\mathrm{B}}\right)^{1 / 2}\)
B \(\frac{\mu_{0} \mathrm{eB}}{4 \pi \mathrm{v}}\)
C \(\left(\frac{\mu_{0} \mathrm{ev}}{4 \pi \mathrm{B}}\right)^{1 / 2}\)
D \(\frac{\mu_{0} \mathrm{ev}}{4 \pi \mathrm{B}}\)

Verified Solution

Answer & Solution

Correct Answer

(C) \(\left(\frac{\mu_{0} \mathrm{ev}}{4 \pi \mathrm{B}}\right)^{1 / 2}\)

Step-by-step Solution

Detailed explanation

The magnetic field at the center of a circular orbit of a moving electron is given by
\(\begin{array}{l}
B=\frac{\mu_{0}}{4 \pi} \times \frac{e}{r^{3}}(v \times r) \\
B=\frac{\mu_{0}}{4 \pi} \times \frac{e v \sin \theta}{r^{2}}
\end{array}\)
As, in a circular path, the angle between radius vector and velocity vector is \(90^{\circ}\).
\(\begin{array}{l}
\mathrm{B}=\frac{\mu_{0}}{4 \pi} \times \frac{\mathrm{ev}}{\mathrm{r}^{2}} \\
\therefore \mathrm{r}=\left(\frac{\mu_{0} \mathrm{ev}}{4 \pi \mathrm{B}}\right)^{1 / 2}
\end{array}\)

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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An electron moves in a circular orbit with uniform speed ' \(v^{\prime}\). It produces amagnetic field 'B' at the centre of the circle. The radius of the circle is\(\left[\mu_{0}=\right.\) permeability of free space, \(\mathrm{e}=\) electronic charge \(]\)

Step-by-step Solution

An electron moves in a circular orbit with uniform speed ' \(v^{\prime}\). It produces a
magnetic field 'B' at the centre of the circle. The radius of the circle is
\(\left[\mu_{0}=\right.\) permeability of free space, \(\mathrm{e}=\) electronic charge \(]\)