Question

The force on a wire is a maximum of 4.50×10−2 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 B

If the pole faces have a diameter of 19.0 cm , estimate the current in the wire if the field is 0.140 T .

Express your answer to three significant figures and include the appropriate units.

Part C

If the wire is tipped so that it makes an angle of 15.0 ∘ with the horizontal, what force will it now feel? [Hint: What length of wire will now be in the field?]

Express your answer to three significant figures and include the appropriate units.

Answer 1

Part B.

Maximum Magnetic force due to a current-carrying wire is given by:

F_max = I*L*B

I = F_max/(L*B)

Given that Force on wire is F_max = 4.50*10^-2 N

L = Length of wire = Diameter of pole face = 19.0 cm = 0.19 m

B = Magnetic field = 0.140 T

So,

I = Current in wire

I = 4.50*10^-2/(0.19*0.140)

I = 1.69 Amp

Part C.

Now when wire is tipped so that it makes 15 deg angle with horizontal, So

F = I*LxB = I*L*B*sin theta

Or we can directly use length of wire which is perpendicular to the magnetic field which will be given by:

Lp = L*sin theta

here theta = angle between current and magnetic field = 90 - 15 deg = 75 deg

So,

F = I*L*B*sin theta

F = 1.69*0.19*0.140*sin 75 deg

F = 4.34*10^-2 N

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