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

Answer:

$1.714\ \text{A}$

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

$F=IBL\sin\theta\\\Rightarrow I=\dfrac{F}{BL\sin\theta}\\\Rightarrow I=\dfrac{0.018}{0.03\times 35\times 10^{-2}\times \sin90^{\circ}}\\\Rightarrow I=1.714\ \text{A}$

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