Question

Consider a 10-cm long piece of a wire which carries a current of 10 A. Find the magnitude of the magnetic field due to the piece at a point which makes an equilateral triangle with the ends of the piece.

Answer 1

Let AB be the wire of length 10 cm and P be the required point.
Given:
Magnitude of current, i = 10 A 
The angles made by points A and B with point P are

\[\theta_1 = 30^\circ \text{ and }  \theta_2 = 30^\circ \] , respectively.

∴ Separation of the point from the wire, d =  \[\frac{\sqrt{3}}{2}a = \frac{\sqrt{3}}{2} \times 10 = 5\sqrt{3} \] cm

Thus, the magnetic field due to current in the wire is given by

 

\[B = \frac{\mu_0 i}{4\pi d}(\sin \theta_1 + \sin \theta_2 )\]
\[ = \frac{{10}^{- 7} \times 10}{5\sqrt{3} \times {10}^{- 2}}\left( \frac{1}{2} + \frac{1}{2} \right)\]
\[ = 11 . 54 \times {10}^{- 6} T\]