2019 Odisha Moving Charges and MagnetismMagnetic Field due to a Circular Coil Medium Question 6 of 15 in Magnetic Field due to a Circular Coil

A straight conductor carrying current i splits into two parts as shown in the figure. The radius of the circular loop is R. The total magnetic field at the centre P of the loop is:

A Zero
B $\frac{3\mu_0 i}{32R}$, outward
C $\frac{3\mu_0 i}{32R}$, inward
D $\frac{\mu_0 i}{2R}$, inward

The answer is A.

Solution

For an arc, $B=\frac{\mu_0 i\theta}{4\pi R}$. The two arcs (270° and 90°) are in parallel, so $\frac{i_1}{i_2}=\frac{R_2}{R_1}=\frac{l_2}{l_1}=\frac{\theta_2}{\theta_1}$, giving $i_1\theta_1=i_2\theta_2$. The fields are equal in magnitude and opposite in direction, so the net field at P is zero.

Concept Magnetic field due to circular arcs

Magnetic Field due to a Circular Coil in past papers

15 questions from this chapter have appeared across 12 exam years.