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An element of 0.05hati m is placed at th...

An element of `0.05hati m` is placed at the origin as shown in figure which carries a large current of 10 A. The magnetic field at a distance of 1 m in perpendicular direction is

A

`4.5xx10^(-8)T`

B

`5.5xx10^(-8)T`

C

`5.0xx10^(-8)T`

D

`7.5xx10^(-8)T`

Text Solution

Verified by Experts

The correct Answer is:
C
`dB=(mu_(0))/(4pi)(Idl sin theta)/(r^(2))`
Here, `dl=Deltax=0.05m, l=10A, r =1 m`
`sin theta=sin 90^(@)=1,`
`therefore" "dB=10^(-7)xx(10xx0.05xx1)/((1)^(2))=0.50xx10^(-7)=5.0xx10^(-8)T`
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