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
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
Show 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.
Updated On: Jul 9, 2026
- \(2\)
- \(4\)
- \(6\)
- \(10\) \bigskip
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The Correct Option is A
Solution and Explanation
Concept:
Magnetic field at the centre of a circular coil of \(N\) turns is
\[
B_{\text{coil}}
=
\frac{\mu_0 N I}{2R}.
\]
Magnetic field due to a long straight conductor at distance \(d\) is
\[
B_{\text{wire}}
=
\frac{\mu_0 I}{2\pi d}.
\]
Since the resultant magnetic field at the centre is zero,
\[
B_{\text{coil}}=B_{\text{wire}}.
\]
Step 1: Calculate the magnetic field due to the circular coil. Given, \[ R=0.1\,\text{m}, \qquad I=2\,\text{A}. \] Therefore, \[ B_{\text{coil}} = \frac{\mu_0 N(2)}{2(0.1)} = 10\mu_0 N. \]
Step 2: Calculate the magnetic field due to the straight wire. Given, \[ I=20\pi\,\text{A}, \qquad d=0.5\,\text{m}. \] Thus, \[ B_{\text{wire}} = \frac{\mu_0(20\pi)} {2\pi(0.5)}. \] \[ B_{\text{wire}} = 20\mu_0. \]
Step 3: Equate the two magnetic fields. \[ 10\mu_0N = 20\mu_0. \] \[ N=2. \]
Step 4: Write the final answer. \[ \boxed{N=2} \] \[ \boxed{\text{Answer = (A)}} \]
Step 1: Calculate the magnetic field due to the circular coil. Given, \[ R=0.1\,\text{m}, \qquad I=2\,\text{A}. \] Therefore, \[ B_{\text{coil}} = \frac{\mu_0 N(2)}{2(0.1)} = 10\mu_0 N. \]
Step 2: Calculate the magnetic field due to the straight wire. Given, \[ I=20\pi\,\text{A}, \qquad d=0.5\,\text{m}. \] Thus, \[ B_{\text{wire}} = \frac{\mu_0(20\pi)} {2\pi(0.5)}. \] \[ B_{\text{wire}} = 20\mu_0. \]
Step 3: Equate the two magnetic fields. \[ 10\mu_0N = 20\mu_0. \] \[ N=2. \]
Step 4: Write the final answer. \[ \boxed{N=2} \] \[ \boxed{\text{Answer = (A)}} \]
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