A proton having an initial velocity of 21.3i Mm/s enters a uniform magnetic fiel

Question

A proton having an initial velocity of 21.3i Mm/s enters a uniform magnetic field of magnitude 0.320 T with a direction perpendicular to the proton's velocity. It leaves the field-filled region with velocity -21.3j Mm/s. (a) Determine the direction of the magnetic field. i -j k -i -k j (b) Determine the radius of curvature of the proton's path while in the field. m (c) Determine the distance the proton traveled in the field. m (d) Determine the time interval for which the proton is in the field.

Explanation / Answer

a) Consider that if the speed changes from positive x-direction to negative y-direction, the proton rotates clockwise. So, when the proton enter to the magnetic field region the force on it is downward or points to the negative y-direction

F = - F j   

then, the direction of the magnetic field must be at the positive z-direction

b) the radius of the curve is

R = ( m v ) / ( q B )

R = ( 1.67x10-27 Kg * 21.3x106 m/s ) / ( 1.60x10-19 C * 0.320 T )

R = 0.694 m

c) the distance runs by the proton us

d = R

where    = / 2 then

d = ( / 2 ) * 0.694 m

d = 1.09 m

d) the time is a quarter of the period

t = T / 4

where,

T = ( 2 m ) / ( q B )

then

t = ( m ) / ( 2 q B )

t = ( * 1.67x10-27 Kg ) / ( 2 * 1.60x10-19 C * 0.320 T )

t = 5.12x10-8 s

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