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A short bar magnet of magnetic moment $5.25 \times 10^{-2} \mathrm{JT}^{-1}$ is placed with its axis perpendicular to the earth's field direction. At what distance from the centre of the magnet, the resultant field is inclined at $45^{\circ}$ with earth's field on (a) its normal bisector and (b) its axis. Magnitude of the earth's field at the place is given to be $0.42 \mathrm{G}$. Ignore the length of the magnet in comparison to the distances involved.
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Verified Answer
Given: Magnetic moment $=5.25 \times 10^{-2} \mathrm{~J} / \mathrm{T}$
$$
\theta=45^{\circ}, \mathrm{H}=0.42 \mathrm{G}=0.42 \times 10^{-4} \mathrm{~T}
$$
To find: Distance at which angle between resultant field and $\mathrm{H}$ is $45^{\circ}$ along
(i) equatorial line (ii) axial line
Formula used:
(i) $\mathrm{B}_{\mathrm{eq}}=\frac{\mu_0}{4 \pi} \frac{\mathrm{M}}{\mathrm{r}^3} \quad$ (ii) $\mathrm{B}_{\text {axial }}=\frac{\mu_0}{4 \pi} \frac{2 \mathrm{M}}{\mathrm{r}^3}$
(i) For the Bar magnet, the normal bisector field and horizontal component of earth's magnetic field will have a resultant at $45^{\circ}$ only when both are equal and perpendicular.
$$
\begin{aligned}
& \text { i.e. } \mathrm{B}_{\mathrm{eq}}=\mathrm{H} \\
\Rightarrow & \mathrm{B}_{\mathrm{eq}}=\frac{\mu_0}{4 \pi} \frac{\mathrm{M}}{\mathrm{r}^3}=\mathrm{H} \\
\Rightarrow \mathrm{r} &=\left(\frac{\mu_0}{4 \pi} \frac{\mathrm{M}}{\mathrm{H}}\right)^{1 / 3} \\
&=\left(\frac{10^{-7} \times 5.25 \times 10^{-2}}{0.42 \times 10^{-4}}\right)^{1 / 3} \\
&=\left(125 \times 10^{-6}\right)^{1 / 3}=5 \times 10^{-2} \mathrm{~m}=5 \mathrm{~cm}
\end{aligned}
$$
Again resultant will be at $45^{\circ}$
when $\mathrm{B}_{\text {axial }}=\mathrm{H} \Rightarrow \mathrm{B}_{\text {axial }}=\frac{\mu_0}{4 \pi} \frac{2 \mathrm{M}}{\mathrm{r}^3}=\mathrm{H}$
$\Rightarrow \mathrm{r}=\left(\frac{\mu_0}{4 \pi} \frac{2 \mathrm{M}}{\mathrm{H}}\right)^{1 / 3}=\left(\frac{10^{-7} \times 5.25 \times 2 \times 10^{-2}}{0.42 \times 10^{-4}}\right)^{1 / 3}$
$=\left(250 \times 10^{-6}\right)^{1 / 3}=6.3 \times 10^{-2}=6.3 \mathrm{~cm}$
$$
\theta=45^{\circ}, \mathrm{H}=0.42 \mathrm{G}=0.42 \times 10^{-4} \mathrm{~T}
$$
To find: Distance at which angle between resultant field and $\mathrm{H}$ is $45^{\circ}$ along
(i) equatorial line (ii) axial line
Formula used:
(i) $\mathrm{B}_{\mathrm{eq}}=\frac{\mu_0}{4 \pi} \frac{\mathrm{M}}{\mathrm{r}^3} \quad$ (ii) $\mathrm{B}_{\text {axial }}=\frac{\mu_0}{4 \pi} \frac{2 \mathrm{M}}{\mathrm{r}^3}$
(i) For the Bar magnet, the normal bisector field and horizontal component of earth's magnetic field will have a resultant at $45^{\circ}$ only when both are equal and perpendicular.
$$
\begin{aligned}
& \text { i.e. } \mathrm{B}_{\mathrm{eq}}=\mathrm{H} \\
\Rightarrow & \mathrm{B}_{\mathrm{eq}}=\frac{\mu_0}{4 \pi} \frac{\mathrm{M}}{\mathrm{r}^3}=\mathrm{H} \\
\Rightarrow \mathrm{r} &=\left(\frac{\mu_0}{4 \pi} \frac{\mathrm{M}}{\mathrm{H}}\right)^{1 / 3} \\
&=\left(\frac{10^{-7} \times 5.25 \times 10^{-2}}{0.42 \times 10^{-4}}\right)^{1 / 3} \\
&=\left(125 \times 10^{-6}\right)^{1 / 3}=5 \times 10^{-2} \mathrm{~m}=5 \mathrm{~cm}
\end{aligned}
$$
Again resultant will be at $45^{\circ}$
when $\mathrm{B}_{\text {axial }}=\mathrm{H} \Rightarrow \mathrm{B}_{\text {axial }}=\frac{\mu_0}{4 \pi} \frac{2 \mathrm{M}}{\mathrm{r}^3}=\mathrm{H}$
$\Rightarrow \mathrm{r}=\left(\frac{\mu_0}{4 \pi} \frac{2 \mathrm{M}}{\mathrm{H}}\right)^{1 / 3}=\left(\frac{10^{-7} \times 5.25 \times 2 \times 10^{-2}}{0.42 \times 10^{-4}}\right)^{1 / 3}$
$=\left(250 \times 10^{-6}\right)^{1 / 3}=6.3 \times 10^{-2}=6.3 \mathrm{~cm}$
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