Question

The force on a wire is a maximum of 7.50×10⁻² N when placed between the pole faces of a magnet. The current flows horizontally to the right and the magnetic field is vertical. The wire is observed to "jump" toward the observer when the current is turned on.

Part A: If the pole faces have a diameter of 12.0 cm , estimate the current in the wire if the field is 0.260 T .

Part B: If the wire is tipped so that it makes an angle of 12.0 ∘ with the horizontal, what force will it now feel? [Hint: What length of wire will now be in the field?]

Answer 1

Part A -

Maximum force, F = 7.50 x 10^-2 N = 0.0750 N

Suppose 'I' is the requisite current in the wire.

Given that -

B = 0.260 T

L = 12 cm = 0.12 m

Use the expression -

F = BIL

=> I = F/BL

Put the values -

I = 0.0750 / (0.260 x 0.12) = 2.40 A

Part B -

In this case -

F(new) =F(old) cos 12
= 0.0750 x cos12 = 0.07336 N = 7.336 x 10^-2 N (Answer)