Question

The horizontal component of earth’s magnetic field at a certain place is 3.0 × 10-5 T and having a direction from the geographic south to geographic north. The force per unit length on a very long straight conductor carrying a steady current of 1.2 A in east to west direction is

• A. 3.0 × 10-5 N m-1
• B. 3.2 × 10-5 N m-1
• C. 3.6 × 10-5 N m-1
• D. 3.8 × 10-5 N m-1

Answer 1

Answer: (C)

3.6 × 10-5 N m-1

Answer 2

Since

Now force per unit length $\mathrm { f } = \frac { \mathrm { F } } { \mathrm { l } } = \mathrm { IB } \sin \theta$

When the current is flowing from east to west then $\theta = 90 ^ { \circ }$ hence

$\mathrm { f } = \mathrm { IB } \sin 90 ^ { \circ } = 1.2 \times 3 \times 10 ^ { - 5 } \times 1 = 3.6 \times 10 ^ { - 5 } \mathrm { Nm } ^ { - 1 }$