Question

A charge \(q\) coulomb is circulating in an orbit of radius \(r\) metres making \(n\) revolutions per second. The magnetic field (in N/A m) produced at the centre of the circle is

• A. \[ \frac{2nq}{nr}\times10^{-7} \]
• B. \[ \frac{2nq}{r}\times10^{-7} \]
• C. \[ \frac{2\pi nq}{r}\times10^{-7} \]
• D. \[ \frac{2\pi rn}{q}\times10^{-7} \]

Hint:

For a charge revolving with frequency \(n\), \[ I=nq \] Always convert the revolving charge into an equivalent current first, then apply \[ B=\frac{\mu_0 I}{2r}. \]

Answer 1

The Correct Option is C

Concept: A revolving charge constitutes an electric current. If a charge \(q\) makes \(n\) revolutions per second, then \[ I=\frac{q}{T} \] Since \[ n=\frac{1}{T} \] therefore \[ I=nq \] The magnetic field at the centre of a circular current loop is \[ B=\frac{\mu_0 I}{2r} \] where \[ \mu_0=4\pi\times10^{-7}\,\text{N A}^{-2} \]

Step 1: Calculate the equivalent current. \[ I=nq \]

Step 2: Substitute into the expression for magnetic field. \[ B=\frac{\mu_0(nq)}{2r} \] Substituting \[ \mu_0=4\pi\times10^{-7} \] \[ B = \frac{4\pi\times10^{-7}\times nq}{2r} \] \[ B = \frac{2\pi nq}{r}\times10^{-7} \]

Step 3: State the answer. \[ \boxed{ B=\frac{2\pi nq}{r}\times10^{-7} } \] Hence, the correct option is \[ \boxed{(C)} \]