Sankalp JEE Full Test-1 Question-45 Solution

Question 45. The magnetic field at the centre of a wire loop formed by two semicircular wires of radii R_{1}=2 \pi ~m and R_{2}=4 \pi ~m carrying current I=4 ~A as per figure given below is \alpha \times 10^{-7} ~T. The value of \alpha is
____ . (Centre O is common for all segments)

Answer (3.00)

Explanation:

\frac{\mu_{0} i}{2 R_{2}}\left(\frac{\pi}{2 \pi}\right) \otimes+\frac{\mu_{0} i}{2 R_{1}}\left(\frac{\pi}{2 \pi}\right) \otimes \left(\frac{\mu_0i}{4R_2}+\frac{\mu_0i}{4R_1}\right)\otimes \frac{4\pi\times10^7\times4}{4\times4\pi}+\frac{4\pi\times10^{-7}\times4}{4\times2\pi} =3\times10^{-7}=\alpha\times10^{-7} \alpha=3