Question

A long, straight wire is turned into a loop of radius $10cm$ (see figure). If a current of $8A$ is passed through the loop, then the value of the magnetic field and its direction at the centre $C$ of the loop shall be close to

• A. $5.0\times 10^{-5}N{A}^{-1}{m}^{-1}$, upwards
• B. $3.4\times 10^{-5}N{A}^{-1}{m}^{-1}$, upwards
• C. $1.6\times 10^{-5}N{A}^{-1}{m}^{-1}$, downwards
• D. $1.6\times 10^{-5}N{A}^{-1}{m}^{-1}$, upwards

Answer 1

Answer: (B)

Magnetic field $B$ at the centre of a coil carrying a current, $I$ is
$B_{\text{coil }}=\frac{{\mu }_{0}I}{2r}$ (upwards)
$B$ due to wire, $B_{\text{wire }}=\frac{{\mu }_{0}I}{2\pi r}$ (downwards)
Magnetic field at centre $C$,
$B_C=B_{\text{coil }}+B_{\text{wire }}$
$=\frac{{\mu }_{0}I}{2r}$ (upward) $+\frac{{\mu }_{0}I}{2\pi r}$ (downwards)
$=\frac{{\mu }_{0}I}{2r}-\frac{{\mu }_{0}I}{2\pi r}=\frac{{\mu }_{0}I}{2r}\left[1-\frac{1}{\pi }\right]$ upwards
Given $I=8A,r=10\times 10^{-2}m$
$=\frac{4\pi \times 10^{-7}\times 8}{2\times 10\times 10^{-2}}\left[1-\frac{1}{3.14}\right]$ upwards
$=3.424\times 10^{-5}$ upwards