Question

a) The force on a wire is a maximum of 8.50×10⁻² 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. If the pole faces have a diameter of 19.0 cm , estimate the current in the wire if the field is 0.460 T . 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?]

b) The force on a wire carrying 7.95 A is a maximum of 2.13 N when placed between the pole faces of a magnet. If the pole faces are 52.5 cm in diameter, what is the approximate strength of the magnetic field?

Answer 1

Answer:

Given, maximum force on the wire is F_max = 8.50 x 10^-2 N, diameter of the pole face is d = 19.0 cm = 0.19 m and magnitude of the magnetic field is B = 0.460 T and angle $\theta$ = 12⁰ with the horizontal so $\phi$ = 90⁰-12⁰ = 78⁰

The direction of the magnetic field is vertical, so the top pole is south and the bottom pole is north.

The maximum magnetic force on the current carrying wire is F_max = ILB

Therefore, the current flow in the wire is I = F_max/LB

Here, the diameter of the pole is equal to the length of the wire placed in the magnetic field B,i.e., d = L = 0.19 m

I = (8.50 x 10^-2 N) / (0.19 m)(0.460 T) = 0.972 A

If the wire is tipped so that it makes an angle of 12⁰ with the horizontal, so the force experienced by the wire is

F = F_max sin$\phi$ = (8.50 x 10^-2 N) sin78⁰ = 8.31 x 10^-2 N

(b) Given, current I = 7.95 A, F_max = 2.13 N and diameter of the pole face is 52.5 cm. The length of the wire placed in the magnetic field is equal to the diameter of the pole face,i.e., L = 52.5 cm = 0.525 m

Therefore, the strength of the magnetic field is B = F_max/IL           [ since F_max = ILB ]

B = (2.13 N) / (7.95 A) (0.525 m) = 0.510 T