Question

The magnetic force on a straight wire 0.69 m long is 1.5 x 10-3 N. The current in the wire is 16.9 A. What is the magnitude of the magnetic field that is perpendicular to the wire

Answer 1

Answer:

Magnitude of magnetic field is 1.29 x 10⁻⁴ T

Explanation:

Given :

Current flowing through the wire, I = 16.9 A

Length of wire. L = 0.69 m

Magnetic force experienced by the wire, F = 1.5 x 10⁻³ N

Consider B be the applied magnetic field.

The relation to determine the magnetic force experienced by current carrying wire is:

F = ILBsinθ

Here θ is the angle between magnetic field and current carrying wire.

According to the problem, the magnetic field and current carrying wire are perpendicular to each other, that means θ = 90⁰. So, the above equation becomes:

F = ILB

$B=\frac{F}{IL}$

Substitute the suitable values in the above equation.

$B=\frac{1.5\times10^{-3} }{16.9\times0.69}$

B = 1.29 x 10⁻⁴ T