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A short bar magnet of magnetic moment 5.25 × 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° with earth’s field on (a) its normal bisector and (b) its axis. The magnitude of the earth’s field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.
A short bar magnet of magnetic moment 5.25 × 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° with earth’s field on (a) its normal bisector and (b) its axis. The magnitude of the earth’s field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.
CBSE | Class 12 | Magnetism and Matter | NCERT | Physics
A short bar magnet of magnetic moment 5.25 × 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° with earth’s field on (a) its normal bisector and (b) its axis. The magnitude of the earth’s field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.

Answer :

Magnetic moment of the bar magnet is given by M = 5.25 × 10–2 J T–1 

Magnitude of earth’s magnetic field at a place is given by H = 0.42 G = 0.42×10-4 T

  • The magnetic field on the ordinary bisector at a distance R from the centre of the magnet is given by –

$B=\frac{{{\mu }_{0}}M}{4\pi {{R}^{3}}}$

Where \[{{\mu }_{0}}=4\pi \times {{10}^{-7}}Tm{{A}^{-1}}\]

It is known that B = H when the resultant field is inclined at 45° with earth’s field. So, we have –

$\therefore \frac{{{\mu }_{0}}M}{4\pi {{R}^{3}}}=H=0.42\times {{10}^{-4}}$

\[{{R}^{3}}=\frac{{{\mu }_{0}}M}{0.42\times {{10}^{-4}}\times 4\pi }=\frac{4\pi \times {{10}^{-7}}\times 5.25\times {{10}^{-2}}}{0.42\times {{10}^{-4}}\times 4\pi }\]

\[{{R}^{3}}=12.5\times {{10}^{-5}}\]

\[\therefore R=5cm\]

  • The magnetic field at a distance ‘R’ from the centre of the magnet on its axis can be given as follows –

${B}’=\frac{{{\mu }_{0}}2M}{4\pi {{d}_{1}}^{3}}=H$

Since, the resultant field is inclined at 45° with earth’s field.

Where, M is the magnetic moment and

\[{{\mu }_{0}}=4\pi \times {{10}^{-7}}Tm{{A}^{-1}}\]

$\therefore {{R}^{3}}=\frac{{{\mu }_{0}}2M}{4\pi \times H}$

\[{{R}^{3}}=\frac{4\pi \times {{10}^{-7}}\times 2\times 5.25\times {{10}^{-2}}}{0.42\times {{10}^{-4}}\times 4\pi }=25\times {{10}^{-5}}\]

\[\therefore R=6.3cm\]

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