The figure below shows a proton moving at velocity = (-200 m/s); toward a long s

Question

The figure below shows a proton moving at velocity = (-200 m/s); toward a long straight wire with current i = 360 mA. At the instant shown, the proton's distance from the wire is d = 3.08 cm. What is the magnetic force on the proton due to the current? (Express your answer in vector form.) A toroid having a square cross section, 3.00 cm on a side, and an inner radius of 11.0 cm has 650 turns and carries a current of 0.700 A. (It is made up of a square solenoid bent into a doughnut shape.) What is the magnitude of the magnetic field inside the toroid at the Inner radius? What is the magnitude of the magnetic field inside the toroid at the outer radius?

Explanation / Answer

Magnetic Field due to Current carrying wire at distance d is given by, B = (uo*I)/(2*pi*d)
Magnetic Force on Proton, F = q* V XB

F = q* v * (uo*I)/(2*pi*d) N
Where,
q = 1.6 * 10^-19 C
v = 200 m/s
d = 3.08 * 10^-2 m
I = 360 * 10^-3 A

F = [(1.6 * 10^-19) * 200 * (4*pi*10^-7) * (360 * 10^-3)] /(2*pi*3.08*10^-2)
F = 7.48 * 10^-23 N

Direction of the Force is towards west which means ( - i^).
Therefore Force , F = 7.48 * 10^-23 ( - i^) N

Please post seperate questions in seperate post.

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