Question

A wire loop formed by joining two semi-circular wires of radii $R_1$ and $R_2$ carries a current as shown in the adjoining diagram. The magnetic induction at the center $O$ is

• A. $\frac{{\mu }_{0}I}{4R_1}$
• B. $\frac{{\mu }_{0}I}{4R_2}$
• C. $\frac{{\mu }_{0}I}{4}(\frac{1}{R_1}-\frac{1}{R_2})$
• D. $\frac{{\mu }_{0}I}{4}(\frac{1}{R_1}+\frac{1}{R_2})$

Answer 1

The correct option is D $\frac{{\mu }_{0}I}{4}(\frac{1}{R_1}+\frac{1}{R_2})$

Here the field due to "SR" and "OQ" will be zero because they pass through point "O" and intersect each other. The field is only present due to wires have radii "$R_1$" and "$R_2$".

so, we know that field due to semi-circular current $B$ = $\frac{\mu i}{4R}$.

for semi-circle have radius $R_1$=$\frac{\mu i}{4R_1}$

for semi-circle have radius $R_2$=$\frac{\mu i}{4R_2}$.

on adding both we find our final result,i.e

$B_{net}$=$\frac{\mu i}{4}\times [\frac{1}{R_1}+\frac{1}{R_2}$].

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