A square shaped wire loop of side L is carrying a current I. What is the magnetic field at the point of intersection of diagonals of the square wire loop?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{2\sqrt{2}\mu_0 I}{\pi L} \)
By the Biot-Savart law, a straight wire carrying current \( I \) produces at a perpendicular distance \( d \) the field
\( \displaystyle B = \frac{\mu_0 I}{4\pi d}\left(\sin\theta_1 + \sin\theta_2\right) \)
where \( \theta_1 \) and \( \theta_2 \) are the angles that the two ends of the wire make with the perpendicular from the point.
The centre of the square is at \( d = L/2 \) from each side, and each end of a side is seen at \( 45^\circ \) from the perpendicular.
\( \displaystyle B_{\mathrm{side}} = \frac{\mu_0 I}{4\pi (L/2)}\left(\sin 45^\circ + \sin 45^\circ\right) = \frac{\mu_0 I}{2\pi L} \times \sqrt{2} = \frac{\sqrt{2}\,\mu_0 I}{2\pi L} \)
All four sides give fields in the same direction at the centre (perpendicular to the plane of the loop), so they add:
\( \displaystyle B = 4 \times \frac{\sqrt{2}\,\mu_0 I}{2\pi L} = \frac{2\sqrt{2}\,\mu_0 I}{\pi L} \)