An electron passes through a point 271 cm from a long straig

An electron passes through a point 2.71 cm from a long straight wire as it moves at 32.1% of the speed of light perpendicularly toward the wire. The wire carries a current of 19.1 A. Find the magnitude of the electron\\\'s acceleration at that point. ( answer in m/s^2)

Solution

magnetic field at the given point, B = mue*I/(2*pi*d)

= 4*pi*10^-7*19.1/(2*pi*0.0271)

= 1.409*10^-4 T

speed of electron, v = 0.32% of c

= 0.32*3*10^8

= 9.6*10^7 m/s

magnetic force on electron, F = q*v*B*sin(90)

m*a = q*v*B (According to Newton\'s second law , F = m*a)

a = q*v*B/m

= 1.6*10^-19*9.6*10^7*1.409*10^-4/(9.1*10^-31)

= 2.38*10^15 m/s^2 <<<<<----------Answer