Question

The ratio of the magnetic fields due to small bar magnet in longitudinal and transverse position is (at equal distance from the magnet)

• A. $1/4$
• B. $1/2$
• C. $1$
• D. $2$

Answer 1

Answer: (D)

$2$

$\text{Given:-}$ Magnetic field in longitudinal $=\frac{{\mu }_0}{4\pi }\frac{2md}{{(d^2-l^2)}^2}$

Magnetic field in Transverse $=\frac{{\mu }_0}{4\pi }\frac{m}{{(d^2+l^2)}^{\frac{3}{2}}}$

$\text{Solution:-}$

$tan\left(A\right)=\frac{{\mu }_0}{4\pi }\frac{2md}{{(d^2-l^2)}^2}$ (longitudinal)

$tan\left(B\right)=\frac{{\mu }_0}{4\pi }\frac{m}{{(d^2+l^2)}^{\frac{3}{2}}}$ (Transverse)

in case of short bar magnet $l^2<<d^2$

$tan\left(A\right)=\frac{{\mu }_0}{4\pi }\frac{2m}{d^3}$ (longitudinal)

$tan\left(B\right)=\frac{{\mu }_0}{4\pi }\frac{m}{d^3}$ (Transverse)

From above equation we get,

$\frac{tan\left(A\right)}{tan\left(B\right)}=\frac{2}{1}$

$\text{Hence the correct option is D}$