1.

A bar magnet has a pole strength of 3.6 A m and magnetie length 8 in. Then, choose the correct option(s).

Answer»

Magnetic field at a point on the axis at a distance of 6 cm from the CENTRE towards the north pole is `8.6 xx 10^(-4)` T.
Direction of flield at an axial point towards the north pole will be towards the magnet.
Magnetic field at a point on the broadside-on POSITION will be `7.7xx10^(-5) T`
The magnetic field at a point in the broadside-on position will be parallel to the magnet.

Solution :The point in question is in end-on position, so the magnetic field is
`B = mu_(0)/(4pi)(2Md)/((d^(2)-l^(2))^(2))`
`10^(-7) = (Tm)/A xx(2xx3.6Am xx0.08m xx 0.06m)/([(0.06)^(2)-(0.04m)^(2)]^(2))`
` = 8.6 xx 10^(-4)T`
The field will be away from the magnet.
when, the point is in broadside-on position so the field is,
`B = mu_(0)/(4pi)M/(d^(2)+l^(2))^(3//2)`
`= 10^(-7) (Tm)/A xx (3.6Am xx 0.08m)/([(0.06 m)^(2)+(0.04m)^(2)]^(3//2))`
`= 7.7 xx 10^(-5) T`
The field will be parallel to the magnet.


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