Question

Traveling at an initial speed of 1.5 x 106 m/s, a proton enters a region of constant magnetic field of magnitude 1.5 T. If the proton's initial velocity vector makes an angle of 30 with the magnetic field, compute the proton's speed 4 s after entering the magnetic field. 0 A. 5.0 x 105 m/s O B. 7.5 x 105 m/s ° C. 1.5 x 106 m/s 0 D. 3.0 x 106 m/s

Answer 1

Answer:

C) $v = 1.5 \times 10^6 m/s$

Explanation:

As we know that force on a moving charge due to magnetic field is given by

$\vec F = q(\vec v \times \vec B)$

so from above formula we can see that

force on moving charge is always perpendicular to the velocity on charge moving in constant magnetic field

So the work done by this force on moving charge is always zero

so this force can change the direction of velocity of charge but it can not change the magnitude of speed of the charge

so at all instants of time the speed will remain the same

so after t = 4 s the speed of the charge is given as

$v = 1.5 \times 10^6 m/s$