Question

Two concentric coils each of radius equal to 2 $\pi$ cm are placed right angles to each other. If 3A and 4A are the currents flowing through the two coils respectively. The magnetic induction (in Wb m-2) at the center of the coils will be

• A. $5\times {10}^{-5}$
• B. $12\times {10}^{-5}$
• C. $7\times {10}^{-5}$
• D. ${10}^{-5}$

Answer 1

The correct option is A

$5\times {10}^{-5}$

$B_1=\frac{{\mu }_0}{4\pi }.\frac{2\pi I}{r}={10}^{-7}\times \frac{2\pi \left(3\right)}{2\pi \times {10}^{-}2}$

$B_1=3\times {10}^{-5},B_2=\left(\frac{{\mu }_0}{4\pi }\right)\frac{2\pi I}{r}=\frac{{10}^{-7}\times 2\pi \left(4\right)}{2\pi \times {10}^{-2}}$

$B_2=4\times {10}^{-5}$

$B_R=\sqrt{B_1^2+B_2^2}$

$B_R=\sqrt{(3\times {10}^{-5}{)}^2+(4\times {10}^{-5}{)}^2}$

$B_R={10}^{-5}\sqrt{9+16}=5\times {10}^{-5}T$