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 centre of the coils will be

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

Answer 1

Answer: (C)

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

As the two concentric coils are placed at right angle to each other, thus the magnetic induction at centre of respective coils are also at right angle. Let say, magnetic induction due to coil 1 points in x direction and that of coil 2 points in y direction.

Given : $r_1=r_2=2\pi$ cm $=0.02\pi$ m $I_1=3A$ $I_2=4A$

Magnetic induction due to coil 1, $B_1=\frac{{\mu }_o{I}_1}{2r_1}$ $\hat{i}$

$\therefore$ $B_1=\frac{4\pi \times {10}^{-7}\times 3}{\left(2\right)\left(0.02\pi \right)}=3\times {10}^{-5}$ $\hat{i}$ $Wb/m^2$

Magnetic induction due to coil 2, $B_2=\frac{{\mu }_o{I}_2}{2r_2}$ $\hat{j}$

$\therefore$ $B_2=\frac{4\pi \times {10}^{-7}\times 4}{\left(2\right)\left(0.02\pi \right)}=4\times {10}^{-5}$ $\hat{j}$ $Wb/m^2$

Thus total magnetic induction $B_T=\sqrt{B_1^2+B_2^2}$

$\therefore$ $B_T=\sqrt{3^2+4^2}\times {10}^{-5}=5\times {10}^{-5}$ $Wb/m^2$