The magnetic field inside a solenoid of circular cross section is given by B? =btk^, where b = 2.1 T/ms . At time t = 0.50 ?s , a proton is inside the solenoid at x = 4.8 cm , y= z=0 , and is moving with velocity v?  = 5.0 j^Mm/s . The center of the solenoid is at x = y = 0.

Find the electromagnetic force on the proton.

Fx, Fy, Fz = ? N

Answer:

Given
B = bt ^k
b = 2.1 T/ms
t = 0.50 µs = 5*10^-7 s
x = 4.8 cm = 4.8*10^-2 m
y = 0
z = 0
v = 5*10^6 ^j m/s

b = 2.1T/ms (1000 ms/1s ) = 2100 T/s

At t = 0.5 µs = 5*10^-7 s, The magnetic field is :
B = (2100 k) *(5*10^-7) = 10.5*10^-4 k T

The magnetic field on the proton is :
F = qvxB
F = (1.6*10^-19)*(5*10^6 j )x(10.5*10^-4 k) = (8 × 10^-13 j )x(10.5*10^-4 k) = 8.4 × 10^-16 i

E2pir = -pi r^2*b
E = -pi*r^2b/(2pir) = -rb/2 = -(4.8*10^-2)2100/2 = -50.2 N/C

E = - 50.2 N/C

F = qE
F = (1.6*10^-19)*(- 50.2 j ) = -8.032 × 10^-18 j N

Total Force on the proton is :

F = ( -8.032 × 10^-18 j N , 8.4 × 10^-16 i N)