2014 AIPMT Moving Charges and MagnetismBiot-Savart Law – Field due to Straight Wires and Arc Medium Question 4 of 5 in Biot-Savart Law – Field due to Straight Wires and Arc

Two identical long conducting wires AOB and COD are placed at right angle to each other, with one above other such that 'O' is their common point for the two. The wires carry $I_1$ and $I_2$ currents respectively. Point 'P' is lying at distance 'd' from 'O' along a direction perpendicular to the plane containing the wires. The magnetic field at the point 'P' will be:

A $\frac{\mu_0}{2\pi d}\left(\frac{I_1}{I_2}\right)$
B $\frac{\mu_0}{2\pi d}(I_1 + I_2)$
C $\frac{\mu_0}{2\pi d}(I_1^2 - I_2^2)$
D $\frac{\mu_0}{2\pi d}(I_1^2 + I_2^2)^{1/2}$

The answer is D.

Solution

The fields of the two wires at P are perpendicular: $B = \sqrt{B_1^2 + B_2^2}$
$B = \sqrt{\left(\frac{\mu_0I_1}{2\pi d}\right)^2 + \left(\frac{\mu_0I_2}{2\pi d}\right)^2} = \frac{\mu_0}{2\pi d}\sqrt{I_1^2 + I_2^2}$

Concept Perpendicular fields

Biot-Savart Law – Field due to Straight Wires and Arc in past papers

5 questions from this chapter have appeared across 5 exam years.