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Question
A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. The earth’s magnetic field at the place is 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null point (i.e., 14 cm) from the centre of the magnet? (At null points, field due to a magnet is equal and opposite to the horizontal component of earth’s magnetic field.)
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Solution
Earth’s magnetic field at the given place, H = 0.36 G
The magnetic field at a distance d, on the axis of the magnet is given as:
`"B"_1 = μ_0/(4pi) (2"M")/"d"^3 ="H"` ............(i)
Where,
μ0 = Permeability of free space
M = Magnetic moment
The magnetic field at the same distance d, on the equatorial line of the magnet is given as:
`"B"_2 = (μ_0"M")/(4pi"d"^3) = "H"/2` ...........[Using equation (i)]
Total magnetic field, B = B1 + B2
= `"H" + "H"/2`
= 0.36 + 0.18
= 0.54 G
Hence, the magnetic field is 0.54 G in the direction of the earth’s magnetic field.
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