A current of 5.0 A is passed through the coil of a galvanometer having 500 turns and each turns has an average area of `3xx10^(-4) m^(2)` if a torque
A current of 5.0 A is passed through the coil of a galvanometer having 500 turns and each turns has an average area of `3xx10^(-4) m^(2)` if a torque of 1.5 N-m is required for this coil carrying same current to set it parallel to a magnetic field calculate the strength of the magnetic field
A. 20T
B. 25T
C. 23T
D. 21T
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Given current i=5.0 A
number of turns N=500
area `A=3xx10^(-4) m^(2)`
torque `tau =1.5 N-m`
The magnetic moment of a current loop
`M=NiA=500xx0.5xx3xx10^(-4)`
`=0.075 Am^(2)`
Also `tau=MxxBor |tau| = MB sin theta`
where `theta` = angle between B and A
Here `theta =90^(@)`
`therefore tau =MB sin 90^(@) , B=(tau)/(M)=(1.5)/(0.075)=20 T`
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