Question

A short bar magnet of magnetic moment $5.25\times {{10}^{-2}}J{{T}^{-1}}$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 the earth’s field on (a) its normal bisector and (b) its axis. The magnitude of the earth’s field at the places is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.

Answer 1

The magnetic field is computed using the formula that relates the magnetic moment and the radius. The distance from the centre of the magnet equals the distance from the centre of the magnet. The magnetic moment of the magnet will be twice at the axis than at the normal bisector.
Formula used:
$B=\dfrac{{{\mu }_{0}}M}{4\pi {{R}^{3}}}$

Complete answer:
From the given information, we have the data as follows.
A short bar magnet of magnetic moment $5.25\times {{10}^{-2}}J{{T}^{-1}}$is placed with its axis perpendicular to the earth’s field direction. The resultant field is inclined at $45{}^\circ$with the earth’s field. The magnitude of the earth’s field at the places is given to be 0.42 G.
The magnetic moment of the bar magnet, $M=5.25\times {{10}^{-2}}J{{T}^{-1}}$
The magnitude of earth’s magnetic field at a place, $H=0.42G=0.42\times {{10}^{-4}}T$
When the resultant field is inclined at $45{}^\circ$with earth’s field, then, $B=H$

& \dfrac{{{\mu }_{0}}M}{4\pi {{R}^{3}}}=H \\\ & \Rightarrow R=\sqrt[3]{\dfrac{{{\mu }_{0}}M}{4\pi H}} \\\ \end{aligned}$$ Substitute the values in the above formula. $$\begin{aligned} & R=\sqrt[3]{\dfrac{4\pi \times {{10}^{-7}}\times 5.25\times {{10}^{-2}}}{4\pi \times 0.42\times {{10}^{-4}}}} \\\ & \Rightarrow R=\sqrt[3]{12.5\times {{10}^{-5}}} \\\ \end{aligned}$$ Therefore, the distance from the centre of the magnet, the resultant field is inclined at $$45{}^\circ $$with earths field on (a) its normal bisector is, $$\begin{aligned} & R=0.05\,m \\\ & \therefore R=5\,cm \\\ \end{aligned}$$ When the resultant field is inclined at $$45{}^\circ $$with earth’s field, then, $$B'=H$$ $$\begin{aligned} & \dfrac{{{\mu }_{0}}2M}{4\pi {{(R')}^{3}}}=H \\\ & \Rightarrow R'=\sqrt[3]{\dfrac{{{\mu }_{0}}2M}{4\pi H}} \\\ \end{aligned}$$ Substitute the values in the above formula. $$\begin{aligned} & R'=\sqrt[3]{\dfrac{4\pi \times {{10}^{-7}}\times 2\times 5.25\times {{10}^{-2}}}{4\pi \times 0.42\times {{10}^{-4}}}} \\\ & \Rightarrow R'=\sqrt[3]{25\times {{10}^{-5}}} \\\ \end{aligned}$$ Therefore, the distance from the centre of the magnet, the resultant field is inclined at $$45{}^\circ $$with earth's field on (a) its axis is, $$\begin{aligned} & R'=0.063\,m \\\ & \therefore R'=6.3\,cm \\\ \end{aligned}$$ $$\therefore $$ The distance from the centre of the magnet, the resultant field is inclined at $$45{}^\circ $$with earth’s field on (a) its normal bisector is 5 cm and (b) its axis is 6.3 cm. **Note:** When the resultant field will be inclined with the earth’s field at a place, then, the magnetic field equals the magnitude of the earth’s magnetic field at that place. At the normal bisector, the value of the magnetic moment will be the same for before and after inclination of the resultant field, but, at the axis, the value of the magnetic moment doubles after the inclination.