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For a short magnetic dipole of magnetic ...

For a short magnetic dipole of magnetic moment 0.5 `Am^(2)`. Find the magnetic field at a point ( a) 1 m on the axis and from the cetre of the magnet ( dipole ) ( b) 1 m on the equatorial line from the centre ( c) at an angle of `60^(@)` w.r. to the line joining the point and the centre of the dipole at distance 1 m.

Text Solution

Verified by Experts

We know that ,
(a) `" " B_(A)=((mu_(o))/(4pi))(2m)/(r^(3))`tesla
= `(10^(-7)xx2xx0.5)/(1^(3))`
`= 1.0 xx10^(-7)`T
(b) `" " B_(B)=((mu_(o))/(4pi))(m)/(r^(3))`tesla
= `(10^(-7)xx0.5)/(1^(3)) = 0.5 xx10^(-7)`T
(c)`" " B = (mu_(o))/(4pi)(m)/(r^(3))sqrt(3cos^(2)theta+1)=(10^(-7)xx0.5)/(1^(3))sqrt(3cos^(2)60^(@)+1)`
`B = (0.5xx10^(-7)xx1.732)/(4) = 0.2165 xx 10^(-7) T ~~ 0.22 xx 10^(7) `T
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Knowledge Check

  • Two short magnetic dipoles each of magnetic moment 10Am^(2) are placed in end on positon 0.1 m apart from their centres the force acting them is

    A
    `0.6xx10^(7)` N
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    `zero `
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    `4xx10^(-4)`N
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  • Two identical magnetic dipole of magnetic moments 1.0 Am^(2) each placed at a separation of 2m with the resultant magnetic field at point midway between the dipole is

    A
    `5xx10^(-7) T`
    B
    `sqrt(5)xx10^(-7) T`
    C
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