Question

The radius of a coil of wire with \(N\) turns is \(0.1\,\text{m}\) and a current of \(2\,\text{A}\) flows in the coil as shown. A long straight wire carrying a current of \(20\pi\,\text{A}\) is located \(0.5\,\text{m}\) from the centre of the coil. The number of turns in the coil if the resultant magnetic field at the centre of the coil is zero is

• A. \(2\)
• B. \(4\)
• C. \(6\)
• D. \(10\) \bigskip

Hint:

Remember the standard results: \[ B_{\text{coil}} = \frac{\mu_0NI}{2R}, \] \[ B_{\text{wire}} = \frac{\mu_0I}{2\pi d}. \] For zero resultant magnetic field, equate the magnitudes of the opposing magnetic fields.

Answer 1

The Correct Option is A

Concept: Zero net magnetic field at centre: \(B_{\text{coil}} = B_{\text{wire}}\). \(B_{\text{coil}} = \mu_0 N I/(2R)\), \(B_{\text{wire}} = \mu_0 I/(2\pi d)\).

Step 1: \(B_{\text{coil}} = \mu_0 N(2)/(2\cdot0.1) = 10\mu_0 N\). \(B_{\text{wire}} = \mu_0(20\pi)/(2\pi\cdot0.5) = 20\mu_0\). Equate: \(10\mu_0 N = 20\mu_0 \Rightarrow N = 2\).

Step 2: Write the final answer. \(\boxed{N=2}\)