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 $(inWb{m}^{-2})$ 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, $I_1=3A$
$I_2=4A$
$R=2\pi cm=2\pi \times 10^{-2}m$
We know that, magnetic field of a coil,
$B=\frac{{\mu }_{0}I}{2R}$
Now, $B_1=\frac{{\mu }_0{I}_1}{2R}=\frac{{\mu }_0}{2}\times \frac{3}{2\pi \times 10^{-2}}$
$=\frac{{\mu }_0}{4\pi }\times \frac{3}{10^{-2}}$
$=10^{-7}\times \frac{3}{10^{-2}}=3\times 10^{-5}T$
Similarly,
$B_2=\frac{{\mu }_0{I}_2}{2R}=\frac{{\mu }_0}{2}\times \frac{4}{2\pi \times 10^{-2}}$
$=\frac{{\mu }_0}{4\pi }\times \frac{4}{10^{-2}}$
$=10^{-7}\times \frac{4}{10^2}=4\times 10^{-5}T$
Now, net magnetic field at centre of a coil,
$B=\sqrt{B_1^2+B_2^2}$
$B=\sqrt{{\left(3\times 10^{-5}\right)}^2+{\left(4\times 10^{-5}\right)}^2}$
$B=\sqrt{10^{-10}(9+16)}$
$B=5\times 10^{-5}T$