Acurrent carrying wire of length 37 cm is positioned perpendicular to a uniform magnetic field. if the current is 3.2 a and it is determined that there is a resultant force of 2.5 n on the wire due to the interaction of the current and field, what is the magnetic field strength?

The magnetic field strength is 2.11 T.

Explanation:

It is given that,

Length of the wire, L = 37 cm = 0.37 m

Current, I = 3.2 A

Force, F = 2.5 N

Magnetic force is given by :

Here,

We know that,

Velocity,

Since,

B = 2.11 T

So, the magnetic field strength is 2.11 T. Hence, this is the required solution.

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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