PLACE STEP BY STEP SOLUTION. ALSO, EXPLAIN THE STEPS. EXPLAIN HOW YOU GOT THE FO
ID: 1496433 • Letter: P
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PLACE STEP BY STEP SOLUTION. ALSO, EXPLAIN THE STEPS.
EXPLAIN HOW YOU GOT THE FORMULAS AND HOW TO SOLVE THEM.
3. (10 pts.) A solenoid has a density of turns of 10 turns/m. The intensity of the current in the solenoid is 1-125 A. (a) Find the intensity of the magnetic field inside the solenoid. A proton is launched inside the solenoid at an initial velocity Vo (2500,-1500, 6000) m/s (the z-axis is parallel to the magnetic field lines) Calculate (b) the radius of the proton's spiral trajectory, (c) the period of revolution and (d) the pitch of the spiral. (mp=1.673x10-27kg, e-1.602x10-19C)Explanation / Answer
a) magnetic field inside a solenoid is given by:
B = n uo i
=10^4*4 pi*10^-7*125
=1.5708 T
b) Radius of trajectory is given by
R = mv/qB
here v is velocity perpendicular to field =sqrt(2500^2+1500^2) = 2915.5 m/s
R = 1.6726219 × 10-27 *2915.5/[1.6*10^-19*1.5708] = 0.0000194 m =1.94 *10^-5 m
c)period, T = 2piR/v = 2 pi*1.94 *10^-5 /2915.5 = 4.18*10^-8 s
d)pitch = Vb T = 6000* 4.18*10^-8 m = 0.0002508 = 2.508 * 10^-4 m
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