Question

If the bar magnet in exercise 5.13 is turned around by 180°, where will the new null points be located?

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Solution

The formula to calculate the magnitude of the magnetic field at a distance d on the axis of the magnet is, B 1 = μ 0 4π ( 2m d 1 3 ) Here, the permeability of the free space is μ 0 , the magnetic moment is m, the magnitude of the magnetic field is B 1 and the distance from the bar magnet is d. If the bar magnet is turned by 180°, then the neutral point will lie on the equatorial line. The formula to calculate the magnitude of the magnetic field at a distance d on the equatorial line of a magnet is, B 2 = μ 0 4π ( m d 2 3 ) The magnetic field in both cases are equal so the formation of the equation is, μ 0 4π ( 2m d 1 3 )= μ 0 4π ( m d 2 3 ) 2 d 1 3 = 1 d 2 3 ( d 2 d 1 ) 3 = 1 2 d 2 = d 1 ( 1 2 ) 1 3 Substituting the values in the above equation, we get: d 2 =( 14 ) ( 1 2 ) 1 3 ≈11.1 cm Thus, the new null point is located at 11.1 cm on the normal bisector.

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