Question

Two identical magnetic dipoles of magnetic moment $2\, A\, m^2$ are placed at a separation of $2\, m$ with their axes perpendicular to each other in air. The resultant magnetic field at a midpoint between the dipoles is

• A. $4 \sqrt{5} \times 10^{-5} \,T$
• B. $2 \sqrt{5} \times 10^{-5} \,T$
• C. $4 \sqrt{5} \times 10^{-7} \,T$
• D. $2 \sqrt{5} \times 10^{-7} \,T$

Answer 1

The Correct Option is D

Let point $P$ be a midpoint between the dipoles. The point $P$ will be in end-on position with respect to one dipole and in broad-side on position with respect to the other.
$\therefore\quad B_{1} = \frac{\mu_{0}}{4\pi} \frac{2m_{1}}{r^{3}_{1}} = \frac{10^{-7}\times2\times 2}{\left(1\right)^{3}} = 4 \times 10^{-7} \,T$
and $B_{2} = \frac{\mu _{0}}{4\pi } \frac{m_{2}}{r^{3}_{2}} = \frac{10^{-7}\times 2}{\left(1\right)^{3}} = 2 \times 10^{-7} \,T$
As $B_{1}$ and $B_{2}$ are perpendicular to each other, therefore the resultant magnetic field at point $P$ is
$B = \sqrt{B^{2}_{1} + B^{2}_{2}} = \sqrt{\left(4 \times 10^{-7}\right)^{2} + \left(2 \times 10^{-7}\right)^{2}}$
$= 10^{-7} \sqrt{16 + 4} = 10^{-7} \sqrt{20} = 2\sqrt{5} \times 10^{-7}\,T$