1) F = 5.78 * 10^-16 N
v = 5.65 * 10^4 m/s
B = 3.2 *10^-2 T
Using, F= qVB
5.78 * 10^-16 = n * 1.6 * 10^-19 * 5.65 * 10^4 * 3.2 * 10^-2
n = 2
Therefore, elementary charges carried by the particle = 2
3) The magnetic field points to the page outside the loop , therefore current is moving in clockwise direction (according to right hand thumb rule).
4) F = qVbSin(theta)
F = 1.6*10^-19 * 2.5 * 10^6 * 0.1*10^-4 * Sin(90-35)
F = 3.6 * 10^ -18 N
10) B = N*I/2r
0.8 * 10^-3 = 50*4*3.14 *10^-7 * I /(2*0.15)
I = 3.82 A
Therefore, current through the loop is 3.82 A
1. A force of 5.78x10-16 N acts on an unknown particle that travels at a 90...
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Set up for problem is given in
the picture. a-c are good. Please help with part d (not in
picture)
d) Next, the current in the solenoid is varied in time as the
function, ?(?) = ?0e−t/τ, where ?0 = 13.0 A
and τ = 5.24 μs. Determine the induced current in the coil at t =
4.00 μs. *Answer is 10.9 A, must show all work
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