## Saturday, September 09, 2006

### Multiple Choice Questions on Magnetism

The following question which appeared in the Kerala Engineering Entrance test paper of 2006 has been popular among question setters for long:
A magnetized wire of magnetic moment M and length L is bent in the form of a semicircle of radius ‘r’. The new magnetic moment is
(a) M (b) M/2π (c) M/π (d) 2M/π (e) zero
The pole strength of the magnet, p = M/L. The pole strength of the magnet is unchanged, but the moment is changed since the poles come closer on bending the wire, thereby changing the magnetic length of the magnet from L to L'
We have L' = 2r = 2L/π so that the new magnetic moment = pL' = (M/L) ×(2L/π) = 2M/π [Option (d)].
Consider now the following M.C.Q.:
In a hydrogen atom the electron is making 6.6×1015 revolutions per second around the nucleus in an orbit of radius 0.528 Å. The equivalent magnetic dipole moment is approximately ( in Am2)
(a)10-10 (b) 10-15 (c) 10-23 (d) 10-25 (e) 10-17

The orbiting electron is equivalent to a circular current loop, whose magnetic dipole moment is given by M = IA where I is the equivalent current and A is the area of the loop. Therefore, M = (q/T)×(πr2) = qf πr2 where q is the electronic charge, T is the orbital period, r is the orbital radius and f is the frequency of revolution of the electron.
Thus M = 1.6×10-19×6.6×1015×π×(5.28×10-10)2 . This yields a value nearly 10-23 [Option (c)].
Let us now consider the following question:
Two short magnets of dipole moments M and 2M are arranged on the table so that the axial line of the weaker magnet and the equatorial line of the stronger magnet are coinciding. If the separation between the magnets is 2d, what is the magnetic flux density midway between these magnets? Ignore the earth’s magnetic field.
(a) μ0M/4πd3 (b) 3μ0M/4πd3 (c) (μ0M/4πd3)
3 (d) (μ0M/4πd3)5 (e) (μ0M/4πd3
)8
At the point midway between the magnets, the axial field (μ0/4π)(2M/d3) of the magnet of moment M and the equatorial field (μ0/4π)(2M/d3) of the magnet of moment 2M are acting at right angles so that the net field there is √2×(μ0/4π)(2M/d3) = μ0M/4πd3)√8. The correct option therefore is (e).

#### 1 comment:

1. Anonymous3:26 PM

the website is awesome it is quite meaningful and useful than the other try to add some more topics