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A short magnet (M=4xx10^(-2)) lying in a...

A short magnet `(M=4xx10^(-2))` lying in a horizontal plane with its north pole points `37^(@)` east of north. Find the net horizontal field a ta point of the magnet of 0.1m away from it `(B_(h)=11muT)(sin37^(@)=3//5,cos37^(@)=4//5)`

A

`17.56 *10^(-6)T` `tan beta=0.14`

B

`10^(-6)T` `tan beta=0.5`

C

`10^(-5)T` `tan beta=0.14`

D

`10^(6)T` `tan beta=0.14`

Text Solution

Verified by Experts

The correct Answer is:
A

Due to magnet at P
`" "B_(1)=(mu_(0))/(4pi).(2Mcostheta)/(r)^(3)`
`" " =(10^(-7)xx2xx4xx10^(_2)xx(4)/(5))/(0.1)^(2)=6.4xx10^(-6)T`
`" "B_(2)=(mu_(0))/(4pi).(Msintheta)/(r^(3))`
`" " =(10^(-7)xx4xx10^(_2)xx(3)/(5))/(0.1)^(2)=2.4xx10^(-6)T`
Since `B_(1) and B_(h)` are in same position
`" " B'=B_(1)+B_(h)=6.4xx10^(-6)+11xx10^(-6)=17.4xx10^(-6)T`

`" " B_(P)=sqrt(B'^(2)+B_(2)^(2))=sqrt((17.4)^(2)+(2.4)^2)xx10^(-6)`
`" " =sqrt(308.52)=10^(-6)T`
`" " tan beta=(B_(2))/(B)=(2.4xx10^(-6))/(17.4xx10^(-6))=0.14`
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