A bar magnet 20 cm in length is placed with its south pole towards geographical north. The neutral points are situated at a distance of 40 cm from centre of the magnet. If horizontal component of earth's field `3.2 xx 10^(-5) Wb//m^2`, then what is the pole strength of a magnet ?

Options

A. 900 ab — amp x cm
B. 90 ab — amp x cm
C. 450 ab — amp x cm
D. 225 ab — amp x cm

Answer

C

Explanation

Given:
`2l = 20 cm rArr 1 = 10 cm = 0.1 m`
`d = 40 cm rArr d = 0.4 m`
When the south pole of a bar magnet is placed towards the north pole of the earth then the neutral point is obtained on axial line.

`B_(H) = B_("arial") = mu_(0)/(4pi)(2M_(d))/(d^(2) - l^(2))^(2)`
`rArr B_(H) = 10^(-7) (0.8 M)/0.0225`
`rArr M = 3.2 xx 10^(-5) xx 0.0225 xx 10^(7)`
Now, `m = M/(2l), [therefore M = m xx 2l]`
` = (3.2 xx 0.0225 xx 10^(2))/0,2 = 45 Am`
`m = 45 Am = 45 xx 1/10 xx 100 cm = 450 ab `- amp cm