Question

A circular coil of radius $5\text{cm}$ carries an electric current $I$ and magnetic field at the centre is $B$. The magnetic field due to the coil at a point $5\text{m}$ from the centre on the axis will be -

• A. $\frac{B}{{10}^6}$
• B. $\frac{B}{{10}^4}$
• C. $\frac{B}{{10}^2}$
• D. $\frac{B}{10}$

Answer 1

The correct option is A $\frac{B}{{10}^6}$


The magnetic field at the centre $\left(O\right)$ of the loop is,

$B_O=B=\frac{{\mu }_{0}I}{2R}....\left(1\right)$

The magnetic field at a point $\text{P}$ on the axis of loop at a distance $r$ from the centre is,

$B_P=\frac{{\mu }_0}{4\pi }\times \frac{2\pi I{R}^2}{(R^2+r^2{)}^{3/2}}$

For the given problem,

$\because R(=5\text{cm})<<r(=5\text{m})\Rightarrow R^2+r^2\approx r^2$

$\therefore B_P=\frac{{\mu }_0}{4\pi }\times \frac{2\pi I{R}^2}{r^3}=\frac{{\mu }_{0}I{R}^2}{2r^3}....\left(2\right)$

On dividing equation $\left(1\right)$ by $\left(2\right)$, we get,

$\frac{B_O}{B_P}=\frac{\frac{{\mu }_{0}I}{2R}}{\frac{{\mu }_{0}I{R}^2}{2r^3}}=\frac{r^3}{R^3}={\left(\frac{r}{R}\right)}^3$

$\Rightarrow \frac{B_O}{B_P}={\left(\frac{5}{5\times {10}^{-2}}\right)}^3={10}^6$

$\Rightarrow B_P=\frac{B_O}{{10}^6}=\frac{B}{{10}^6}$

Hence, option $\left(A\right)$ is the correct answer.