Question

The horizontal component of the Earth's magnetic field at a certain place is 3.0 × 10^-5 T and the direction of the field is from the geographic South to geographic north . A very long straight conductor is carrying a steady current 1 A . What is the force per unit length on it when it is placed on the horizontal table and the direction of the current is ( a ) east to west ( b ) South to North ??​

Answer 1

Answer:

Explanation:

As, F=I  l  ×  B

F=IIBsinθ

The force per unit length is,

f=  f/l   =IBsinθ

When the current is flowing from east to west,

θ=90  

hence,

f=IB=1×3×10  −5  =3×10  −5  Nm  −1