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Above the ground, a 1 m long wire is suspended horizontally perpendicular to a line running from its center to a nearby observer. A uniform magnetic field of 1.8 Tesla permeates the volume occupied by this wire in the upward direction.

a) Ifa current of 5x10-2 A is passed thru the wire, what is the force on the wire?
b) What is its direction?

Respuesta :

abidemiokin

Answer:

a) 0.09N b) positive x direction

Explanation:

Force on a conductor carrying current in a magnetic field can be expressed as;

F = BILsin(theta) where

F is the force on the conductor (wire)

B is the uniform magnetic field I'm Tesla = 1.8Tesla

I is the current in the wire = 5×10^-2A

L is the length of the wire = 1m

theta is the angle that the conductor make with the magnetic field = 90° (since the wire in the horizontal direction is perpendicular to the field acting upwards)

Substituting this value in the formula to get F we have;

F = 1.8×5×10^-2×1 × sin90°

F = 0.09N

The force on the wire is 0.09N

b) The direction of the force is in the positive x direction since the wire acts horizontally to the magnetic field.

Olajidey

Answer:

F = 9 × 10⁻² N

Explanation:

Given that,

F is the force acting on the wire

B is the uniform magnetic field = 1.8Tesla

I is the current  = 5×10⁻²A

L is the length  = 1m

θ = 90°

F = BILsinθ

F = 5 × 10⁻² × 1 × 1.8

F = 9 × 10⁻² N

b)

The direction of the force along the z-direction is perpendicular to the magnetic field