1.

A rectrangular coil 20cmxx20cm has 100 turns and carries a current of 1 A. It is placed in a uniform magnetic field B = 0.5T with the direction of magnetic field parallel to the plane of the coil. Find the magnitude of the torque required to hold this coil in this position.

Answer»

Zero
200 Nm
2 Nm
10 Nm

Solution :`N=100,I=1A,B=5xx10^(-1)T`
`A=20xx20xx10^(-4)m,theta=90^(@)`
DEFLECTING torque `|vectau|=NIABsintheta`
`|vectau|=100xx1xx400xx10^(-4)xx5xx10^(-1)xxsin90^(@)`
= 2 Nm


Discussion

No Comment Found

Related InterviewSolutions