The moving coil galvanometer has 60 turns, a width of 2 cm and a length of 3 cm. It hangs in uniform magnetic field of 500 cgs unit. If the controlling couple, due to the twist in the suspension wire is 18 dyne cm, then the current flowing through it is
A
`10^2 A`
B
`10^-3 A`
C
`10^-4 A`
D
`3xx10^-5 A`
Text Solution
Verified by Experts
The correct Answer is:
B
`i=(K)/(n AB)theta(because K, theta , n & A" are constant")` `therefore I prop 1//B rArr I_1 B_1=I_2B_2` `therefore I_2=(B_1)/(B_2)I_1=(4xx10^-3xx2)/(8xx10^-3)=1A`
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