How much current is flowing in a wire 4.55 m long if the maximum force on it is 0.575 N when placed in a uniform 0.0700-T field?

1.81A

The magnitude of the magnetic force (F) acting on a current carrying wire is given by;

F = I x L x B sin θ      (i)

Where;

I = current on the wire

B = magnetic strength

L = length of the wire

θ = angle between the wire and the magnetic field

From the question, there is a maximum force (F = 0.575N) of on the wire. This means that the angle between the wire and the magnetic field is 90°.

Also;

the magnetic field(B) = 0.0700-T

the length(L) of the wire = 4.55m

Substitute these values into equation (i) as follows;

0.575 = I x 4.55 x 0.0700 x sin 90°

0.575 = I x 4.55 x 0.0700 x 1

0.575 = I x 0.3185

Solve for I;

I = 0.575 / 0.3185

I = 1.81A

Therefore, the current flowing in the wire is 1.81A

96.77 amp

explanation:

rl = 250/100 = 2.5 ohms.

egl = 250 + (100 × 0.1) + 2 = 262 v.

at 700 r.p.m., eg2 = 262 × 700/1000 = 183.4 v.

if ia is the new current, then eg2 - 2 - (ia × 0.1) = 2.5 ia

which means this will give ia = 96.77 amp.