Question

A current carrying wire is in the shape of a semicircle of radius $R$ and has current $I$. $M$ is midpoint of the arc and point $P$ lies on extension of $MC$ at a distance $2R$ from $M$. Find the magnetic field due to circular arc at point $P$.

• A. $\frac{\mu_0 I}{4R}$
• B. $\frac{\mu_0 I}{8R}$
• C. $\frac{\mu_0 I}{2R}$
• D. $\frac{\mu_0 I}{\pi R}$

Answer 1

Answer: (B)

$\frac{\mu_0 I}{8R}$

Answer 2

The magnetic field at the center of curvature (point C) due to an arc of angle $\theta$ is given by:

$B_{centre} = \frac{\mu_0 I \theta}{4\pi R}$

For a semicircular arc ($\theta = \pi$):

$B_{centre} = \frac{\mu_0 I}{4R}$

When the field-point lies on the current (here at an end of the arc) the "singular" element is taken as contributing one-half its value. Thus,

$B(P) = \frac{1}{2} \cdot \frac{\mu_0 I}{4R} = \frac{\mu_0 I}{8R}$

The magnetic field at point P is $\frac{\mu_0 I}{8R}$, with the direction given by the right-hand rule. For a counter-clockwise current, the field at P is normal to the plane of the arc.