Question

A conducting rod along the equator is 1 m long and carries a current of 15 A from east to west. The magnitude of Earth's magnetic field at the equator is `4/3 xx 10^(-4)` T. The magnitude and direction of the force on the rod are ______.

Options

• 11.25 x 10^-4 N, downward

• 20 x 10^-4 N, downward

• 11.25 x 10^-4 N, upward

• 20 x 10^-4 N, upward.

Answer 1

A conducting rod along the equator is 1 m long and carries a current of 15 A from east to west. The magnitude of Earth's magnetic field at the equator is `4/3 xx 10^(-4)` T. The magnitude and direction of the force on the rod are 20 x 10^-4 N, downward.

Explanation:

F = `|\overset{→}{Il} xx \overset{→}{B_h}|`

= `IlB = (15)(1)(4/3 xx 10^(-4))`

= 20 × 10^-4N

Since `\overset{→}{l}` is toward the east and  `\overset{→}{B_h}` is toward north, `\overset{→}{l} xx  \overset{→}{B_h}` is downward. So, `\overset{→}{F}` is downward.