Question

A vertical straight wire 35.0 cmcm in length carries a current. You do not know either the magnitude of the current or whether the current is moving upward or downward. If there is a uniform horizontal magnetic field of 0.0300 TT that points due north, the wire experiences a horizontal magnetic force to the west of 0.0180 NN. Find the magnitude of the current.

Answer 1

`1.714\ \text{A}`

Step-by-step explanation:

F = Magnetic force = 0.018 N

B = Magnetic field = 0.03 T

L = Length of wire = 35 cm

`\theta` = Angle between current and magnetic field =
`90^(\circ)`

Magnetic force is given by

`F=IBL\sin\theta\\\Rightarrow I=(F)/(BL\sin\theta)\\\Rightarrow I=(0.018)/(0.03* 35* 10^(-2)* \sin90^(\circ))\\\Rightarrow I=1.714\ \text{A}`

The magnitude of the current is
`1.714\ \text{A}`.