Q. Two concentric coils each of radius equal to $2\pi$cm are placed at right angles to each other. 3 A and 4 A are the currents flowing in each coil respectively. The magnetic induction in ${Wbm}^{- 2}$ at the centre of the coils will be $\left( \mu_{0} = 4\pi \times 10^{- 7}\ {WbAm}^{- 1} \right)$
A) $12 \times 10^{- 5}$
B) $10^{- 5}$
C) $5 \times 10^{- 5}$
D) $7 \times 10^{- 5}$
Explanation
(c) $B_{P} = \frac{\mu_{0}I_{2}}{2R}$
$\ \ \ \ \ \ \ = \frac{4\pi \times 10^{- 7} \times 4}{2 \times 0.02\pi} = 4 \times 10^{- 5}\ {Wbm}^{- 2}$
$B_{Q} = \frac{\mu_{0}I_{1}}{2R}$
$\ \ \ \ \ \ \ = \frac{4\pi \times 10^{- 7} \times 3}{2 \times 0.02\pi} = 3 \times 10^{- 5}\ {Wbm}^{- 2}$
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$\therefore B = \ \sqrt{B_{P}^{2} + B_{Q}^{2}}$
$\ \ \ \ \ \ \ \ = \sqrt{\left( 4 \times 10^{- 5} \right)^{2} + \left( 3 \times 10^{- 5} \right)^{2}}$
$\ \ \ \ \ \ \ \ = 5 \times 10^{- 5}\ {Wbm}^{- 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.