Question

A proton experiences no magnetic force if it travels in the direction of 30 degrees North of East.
It will experience 8x10^-17 N magnetic force directed downward into the ground if it travels due east at 1000 m/s.

Find the magnitude and direction of the magnetic field.

Answer 1

F = q*(vB)

given , q = charge on proton = 1.6*10^-19 C

F= 8*10^-17 N ( downward into ground)

v = 1000 m/s ( in east)

vB = v*B sin

where, = angle between v and B

force can be zero only when  vB will be zero, so  sin should be zero

so when,   =0 or 180 , force =0

it is given , force =0, when v is 30 degree north of east,

This shows either , B is also in this direction or 180 to this direction

applying, right hand thumb rule, we get B is in direction 180 to the (30degree)north of east

{because if B is in 30 degree north of east, force will be in upward direction)

i.e 30 degree west of south

now, when velocity is in east direction

angle between v and B = 150

sin150 =0.5

hence,

F= q*v*B*sin(150)

8*10^-17 = 1.6*10^-19 *1000 *B *0.5

B = 1 T

hence, magnetic field is 1 T in the direction 30 west of south