Question

Two concentric coils each of radius equal to $2\pi$ $cm$ are placed at right angles to each other. $3A$ 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. ${10}^{-5}$
• B. $12\times {10}^{-5}$
• C. $7\times {10}^{-5}$
• D. $5\times {10}^{-5}$

Answer 1

The correct option is D $5\times {10}^{-5}$

The magnetic field at the center of a current carrying circular loop of radius $R$ is: $B_{center}=\frac{{\mu }_{0}i}{2R}$
Magnetic field is perpendicular to the plane of the circular loop.
Since the loops are mutually perpendicular to each other, magnetic field produced by both will be perpendicular to each other also.
$B_1=\frac{{\mu }_0{i}_1}{2R}=\frac{{\mu }_0\times 3}{2\times 2\pi \times {10}^{-2}}=\frac{{\mu }_0}{4\pi }\times 3=3\times {10}^{-5}$
$B_2=\frac{{\mu }_0{i}_2}{2R}=\frac{{\mu }_0\times 4}{2\times 2\pi \times {10}^{-2}}=\frac{{\mu }_0}{4\pi }\times 4=4\times {10}^{-5}$
Since $B_1$ and $B_2$ are mutually perpendicular to each other,
$B=\sqrt{B_1^2+B_2^2}={10}^{-5}\sqrt{3^2+4^2}=5\times {10}^{-5}W/m^2$