Q. Two identical short bar magnets, each having magnetic moment of $10\ Am^{2}$, are arranged such that their axial lines are perpendicular to each other and their centres be along the same straight line in a horizontal plane. If the distance between their centres is $0.2\ m$, the resultant magnetic induction at a point midway between them is $(\mu_{0} = 4\pi \times 10^{- 7}Hm^{- 1})$
A) $\sqrt{2} \times 10^{- 7}tesla$
B) $\sqrt{5} \times 10^{- 7}tesla$
C) $\sqrt{2} \times 10^{- 3}tesla$
D) $\sqrt{5} \times 10^{- 3}tesla$
Explanation
(d) [image:5_image55.emf]
From figure $B_{net} = \sqrt{B_{a}^{\ \ 2} + B_{e}^{\ \ 2}}$
$= \sqrt{\left( \frac{\mu_{0}}{4\pi}.\frac{2M}{d^{3}} \right)^{2} + \left( \frac{\mu_{0}}{4\pi}.\frac{M}{d^{3}} \right)^{2}}$
$= \sqrt{5}.\frac{\mu_{0}}{4\pi}.\frac{M}{d^{3}} = \sqrt{5} \times 10^{- 7} \times \frac{10}{(0.1)^{3}} = \sqrt{5} \times 10^{- 3}tesla$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.