Class 12 Physics
CBSE and UP Board
Magnetism and Matter
Chapter-5 Exercise 5.15
NCERT Solutions for Class 12 Physics Chapter 5 Question 15
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 45º with earth’s field on (a) its normal bisector and (b) its axis. 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.
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Magnetic moment of the bar magnet, M = 5.25 x 10⁻2 J T⁻1
Magnitude of earth’s magnetic field at a place, H = 0.42 G = 0.42 x 10⁻4 T
Ans (a).
The magnetic field at a distance R from the centre of the magnet on the normal bisector is given by the relation:
B₁ = μ0 /4π x [M/(R)³]
Where,
μ0 = Permeability of free space = 4π x 10-7 Tm A-1
When the resultant field is inclined at 45° with earth’s field, B = H
Therefore , μ0 /4π x [M/(R)³] =H = 0.42 x 10⁻4
=> R³ = μ0M /( 0.42 x 10⁻4 x 4π )
= (4π x 10-7 x 5.25 x 10⁻2 )/ ( 0.42 x 10⁻4 x 4π ) = 12.5 x 10⁻⁵
Therefore R = 0.05m = 5 cm
Ans (b).
The magnetic field at a distance R1 from the centre of the magnet on its axis is given as:
B¹ = μ0 /4π x [2M/(R¹)³]
The resultant field is inclined at 45° with earth’s field.
Therefore ,
B¹ =H
μ0 /4π x [2M/(R¹)³] = H
(R¹)³ = μ0 2M/ 4πH
= (4π x 10-7 x 2 x 5.25 x 10⁻2)/ (4π x 0.42 x 10⁻4 )
= 25 x 10⁻⁵
Therefore , R¹ = 0.063 m = 6.3 cm