Question

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

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

Answer 1

The Correct Option is C

Given, $I_{1} =3\, A $ $I_{2} =4 A$ $R =2 \pi cm =2 \pi \times 10^{-2}\, m$ We know that, magnetic field of a coil, $B=\frac{\mu_{0} I}{2 R}$ Now, $B_{1} =\frac{\mu_{0} I_{1}}{2 R}=\frac{\mu_{0}}{2} \times \frac{3}{2 \pi \times 10^{-2}} $ $=\frac{\mu_{0}}{4 \pi} \times \frac{3}{10^{-2}} $ $=10^{-7} \times \frac{3}{10^{-2}}=3 \times 10^{-5}\, T$ Similarly, $B_{2} =\frac{\mu_{0} I_{2}}{2 R}=\frac{\mu_{0}}{2} \times \frac{4}{2 \pi \times 10^{-2}}$ $=\frac{\mu_{0}}{4 \pi} \times \frac{4}{10^{-2}}$ $=10^{-7} \times \frac{4}{10^{2}}=4 \times 10^{-5} \,T$ Now, net magnetic field at centre of a coil, $B=\sqrt{B_{1}^{2}+B_{2}^{2}} $ $B=\sqrt{\left(3 \times 10^{-5}\right)^{2}+\left(4 \times 10^{-5}\right)^{2}} $ $B=\sqrt{10^{-10}(9+16)}$ $B=5 \times 10^{-5}\, T$