Question

A short magnet makes 40 oscillations per minute when used in an oscillations magnetometer at a place where the earth's horizontal magnetic field is 25 $\mu T$. Another short magnet of magnetic moment 1.6 A $m^2$ is placed 20 cm east of the oscillation if the magnet has its north pole (a) towards north and (b) towards south.

Answer 1

Here, ${\gamma }_1$ = 40 oscillations/minute,

$B_H=25\mu T$

M of second magnet

= 1.6 A-$m^2$

d= 20 cm = 0.2 m

(a) For north pole facing

${\gamma }_1=\frac{1}{2\pi }\sqrt{\frac{M{B}_H}{1}}$

${\gamma }_2=\frac{1}{2\pi }\sqrt{\frac{M(B_H-B)}{1}}$

B= $\frac{{\mu }_0}{4\pi }\frac{m}{d^3}$

=$\frac{{10}^{-7}\times 1.6}{8\times {10}^{-3}}=20\mu T$

$\frac{{\gamma }_1}{{\gamma }_2}=\sqrt{\frac{B_H}{B_H-B}}$

$\Rightarrow \frac{40}{{\gamma }_2}=\sqrt{\frac{25}{5}}$

$\Rightarrow \frac{40}{{\gamma }_2}=\sqrt{5}$

$\Rightarrow {\gamma }_2=\frac{40}{\sqrt{5}}$ = 17.88

= 18 osci/m.

(b) For north pole facing south

${\gamma }_1=\frac{1}{2\pi }\sqrt{\frac{M{B}_H}{1}}$

${\gamma }_1=\frac{1}{2\pi }\sqrt{\frac{M(B+B_H)}{1}}$

$\frac{{\gamma }_1}{{\gamma }_2}=\frac{B_H}{B+B_H}$

$\Rightarrow \frac{40}{{\gamma }_2}=\sqrt{\frac{25}{45}}$

${\gamma }_2=\frac{40}{\sqrt{\frac{25}{45}}}$ = 54 osci/m.