A wire carries a steady current of 2.20 A. A straight section of the wire is 0.750 m long and lies along the y axis within a uniform magnetic field, vector B = 1.70 k T. If the current is in the +y direction, what is the magnetic force on the section of wire?

2.805 N

Explanation:

Given,

Current, I = 2.20 A

Length of the section,L = 0.750 m

Magnetic field, B = 1.70 T

θ = 90°

Magnetic Force,F = ?

We know,

F = IBL sin θ

F = 2.20 x 1.70 x 0.750 x sin 90°

F = 2.805 N

Hence, magnetic force is equal to 2.805 N.

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friction

explanation:

they us to move forward and backwards