5.15  (a). A short bar magnet of magnetic movement 5.25 \times 10 ^{-2} 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° with earth’s field on  its normal bisector 

Answers (1)

Given,

The magnetic moment of the bar magnet, m = 5.25 \times 10^{-2}\ JT^{-1}

The magnitude of earth’s magnetic field at a place, H = 0.42 G = 0.42 \times 10^{-4}   T

The magnetic field at a distance R from the centre of the magnet on the normal bisector is:

B = \mu_{0}m/4\pi R^3

When the resultant field is inclined at 45° with earth’s field, B = H 

B = H = 0.42\times 10^{-4} T

R^3 = \mu_{0}m/4\pi B  =  4\pi\times10^{-7}\times5.25\times10^{-2}/4\pi\times0.42\times10^{-4}

= 12.5 \times 10^{-5 }m^3

Therefore,  R = 0.05 m = 5 cm

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