2026 Re-NEET Moving Charges and MagnetismMagnetic Field due to a Circular Coil Medium Question 2 of 15 in Magnetic Field due to a Circular Coil

A current $I_0$ flows through a metallic circular loop of radius r as shown in the figure. Resistance of the segment ABC is half that of ADC. Magnitude of magnetic field at the centre O of the loop is:

A $\frac{\mu_0I_0}{12r}$
B $\frac{\mu_0I_0}{4r}$
C $\frac{\mu_0I_0}{2r}$
D $\frac{\mu_0I_0}{2\pi r}$

The answer is A.

Solution

$R_1:R_2=1:2$ and $I\propto\frac{1}{R}$, so $I_1=\frac{2}{3}I_0$ (through ABC) and $I_2=\frac{1}{3}I_0$ (through ADC). The two semicircles give opposite fields at O: $B_{net}=\frac{\mu_0I_1}{4r}-\frac{\mu_0I_2}{4r}=\frac{\mu_0}{4r}\left(\frac{2I_0}{3}-\frac{I_0}{3}\right)=\frac{\mu_0I_0}{12r}$.

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.