At a place where the magnitude of the earth's magnetic field is $4 \times 10^{-5} \mathrm{~T}$, a short bar…

At a place where the magnitude of the earth's magnetic field is $4 \times 10^{-5} \mathrm{~T}$, a short bar magnet is placed with its axis perpendicular to the earth's magnetic field direction. If the resultant magnetic field at a point at a distance of 40 cm from the centre of the magnet on the normal bisector of the magnet is inclined at $45^{\circ}$ with the earth's field, then the magnetic moment of the magnet is
  1. $38.4 \mathrm{Am}^2$
  2. $51.2 \mathrm{Am}^2$
  3. 12.8 Am ${ }^2$
  4. $25.6 \mathrm{Am}^2$

Solution

$\mathrm{B}_{\mathrm{e}}=4 \times 10^{-5} \mathrm{~T}, \mathrm{r}=40 \mathrm{~cm}$
Magnetic field at a point on the normal bisector of the magnetic is $\begin{aligned} & \mathrm{B}=\frac{\mu_0}{4 \pi} \cdot \frac{\mathrm{M}}{\mathrm{r}^3} \\ & \therefore \tan 45^{\circ}=\frac{\mathrm{B}_{\mathrm{e}}}{\mathrm{~B}} \Rightarrow \mathrm{~B}_{\mathrm{e}}=\mathrm{B} \\ & \Rightarrow \mathrm{~B}_{\mathrm{e}}=\frac{\mu_0}{4 \pi} \cdot \frac{\mathrm{M}}{\mathrm{r}^3} \Rightarrow \mathrm{M}=\frac{\mathrm{B}_{\mathrm{e}^3} \mathrm{r}^3}{\frac{\mu_0}{4 \pi}} \\ & \therefore \mathrm{M}=\frac{4 \times 10^{-5} \times(0.4)^3}{10^{-7}}=25.6 \mathrm{Am}^2 \end{aligned}$

Asked in: AP EAMCET 2024 (20 May Shift 2)

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