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

Answer: (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}} $
$B=\sqrt{10^{-10}(9+16)}$
$B=5 \times 10^{-5}\, T$