✔ 最佳答案
1. By right-hand-grip rule, the magnetic field at B should point into the paper.
Magnetic field due to the two long straight wires, B'
= 2 X u0I/4pir = u0I/2pir
Magnetic field due to the semi-circular arc, B"
= u0I/2r
So, the magnetic field at B = B' + B" = (4pi X 10^-7)(5.6)/2(0.049) [1 + 1/pi]
= 9.5 X 10^-5 T
Your answer is wrong.
2. By energy conservation, the electric potential energy loss = Gain in K.E.
So, K.E., 1/2 mu^2 = qV
Speed, u = sqrt(2qV/m)
b. After entering the magnetic field, the speed of the particle is unchanged.
it will experience a magnetic force, which accounts for the centripetal force.
mu^2 / R = quB0
m/q = RB0/u = RB0/sqrt(2qV/m)
sqrt(m/q) = RB0/sqrt(2V)
So, m/q = R^2B0^2 / 2V