Question

A bar magnet of length $14\mathrm{cm}$ is placed in the magnetic meridian with its north pole pointing towards the geographic north pole. A neutral point is obtained at a distance of $18\mathrm{cm}$ from the center of the magnet. If ${\mathrm{B}}_{\mathrm{H}}=0.4\mathrm{G}$, the magnetic moment of the magnet is $(1\mathrm{G}={10}^{-4}\mathrm{T})$

• A. $28.80{\mathrm{JT}}^{-1}$
• B. $2.880{\mathrm{JT}}^{-1}$
• C. $2.880\times {10}^3{\mathrm{JT}}^{-1}$
• D. $2.880\times {10}^2\mathrm{J}{\mathrm{T}}^{-1}$

Answer 1

The correct option is B

$2.880{\mathrm{JT}}^{-1}$

Step 1: Given data

Here we have,

Length of bar magnet, $2\mathrm{l}=14\mathrm{cm}$

$$\begin{gathered} \Rightarrow \mathrm{l}=\frac{14}{2} \\ \Rightarrow \mathrm{l}=7\mathrm{cm} \\ \Rightarrow \mathrm{l}=0.07\mathrm{m} \end{gathered}$$

Magnetic intensity, $\mathrm{H}=0.4\mathrm{G}=0.4\times {10}^{-4}\mathrm{T}$

Distance of a neutral point from the center of magnet, $\mathrm{d}=18\mathrm{cm}=0.18\mathrm{m}$

Step 2: Calculating the value of magnetic moment

We know that magnetic intensity is given by:

$$\begin{gathered} \mathrm{H}=\frac{{\mathrm{\mu }}_0\mathrm{M}}{4\mathrm{\pi }{\left({\mathrm{d}}^2+{\mathrm{l}}^2\right)}^{{\displaystyle \frac{3}{2}}}} \\ \Rightarrow \mathrm{M}=\frac{4\mathrm{\pi }}{{\mathrm{\mu }}_0}{\left({\mathrm{d}}^2+{\mathrm{l}}^2\right)}^{\frac{3}{2}}\mathrm{H} \end{gathered}$$

Here we have,

Magnetic moment$=\mathrm{M}$

$\frac{4\mathrm{\pi }}{{\mathrm{\mu }}_0}={10}^7$

Put the given values in above equation we get,

$$\begin{gathered} \mathrm{M}={10}^7\times {\left({\left(0.18\right)}^2+{\left(0.07\right)}^2\right)}^{\frac{3}{2}}\times 0.4\times {10}^{-4} \\ \Rightarrow \mathrm{M}=2.880{\mathrm{JT}}^{-1} \end{gathered}$$

Therefore the value of magnetic moment is equal to $2.880{\mathrm{JT}}^{-1}$

Hence, option (B) is the correct answer.