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A bar magnet has a pole strength of 15 A m and magnetic length 20 cm. What is the magnetic induction produced by it at a point at a distance of 30 cm from either pole? (mu_(0)/(4pi) = 10^(-7)Wb//Am)

Answer»

`1.11xx10^(-5) Wb//m^(2)`
`0.8xx10^(-4) Wb//m^(2)`
`1.5xx10^(-5) Wb//m^(2)`
`2.75xx10^(-5) Wb//m^(2)`

Solution :`M = m xx 2l = 15 xx 0.2 = 3 A m^(2)`
The point is at a distance of 30 cm from EITHER pole. Hence it must be on the EQUATOR.
`B_(eq) = (mu_(0))/(4pi)M/((d^(2)+l^(2))^(3//2))`
But `(d^(2) + l^(2))^(1//2) = 30cm = 0.3m `
`therefore B_(eq ) =10^(-7) xx 3/(0.3)^(3)=(3xx10^(-7))/(27xx10^(-3))=(10^(-4))/9`
`IMPLIES B = 1.11 xx10^(-5) Wb//m^(2)`


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