Question

A circular coil of radius $R$ carries an electric current. The magnetic field due to the coil at a point on the axis of the coil located at distance $x(x>>>R)$ is measured. If the current in coil is halved and distance of the point at axis is doubled, the new magnetic field will be:

• A. $16$ times of initial value
• B. $\frac{1}{16}$ times of initial value
• C. Doubled
• D. $\frac{1}{8}$ times of initial value

Answer 1

The correct option is B $\frac{1}{16}$ times of initial value

The magnetic field on the axis of coil is given by

$B_{axis}=\frac{{\mu }_{0}I{R}^2}{2(R^2+x^2{)}^{3/2}}$

For $(x>>R)$,

$B_{axis}=\frac{{\mu }_{0}I{R}^2}{2x^3}$

keeping ${\mu }_0$ & $R$ as the constant parameters, we get

$B_{axis}\propto \frac{I}{x^3}$

$\therefore \frac{B_1}{B_2}=\frac{I_1{x}_2^3}{I_2{x}_1^3}=\left(\frac{I_1}{I_2}\right){\left(\frac{x_2}{x_1}\right)}^3$

Given that $I_2=\frac{I_1}{2}$ and $x_2=2x_1)$

$\frac{B_1}{B_2}=\left(2\right){\left[\frac{2x_1}{x_1}\right]}^3=16$

$\therefore B_2=\frac{B_1}{16}$

Thus, magnetic field will reduce by a factor $\frac{1}{16}$.