Question

An electron moving with a speed of 8.8 x105 m/s in the positive z direction experiences zero magnetic force. When it moves in the positive x direction it experiences a force of 2.4 x10-13 N that points in the negative y direction. What is the magnitude and direction of the magnetic field? Magnitude: B= _____________T

Direction: Magnetic field is pointed in ____________(indicate positive or negative)

direction of ______________(indicate which axis i.e. x, y, or z) axis.

Answer 1

Given that,

The magnetic force is,F=2.4 x10-13 N

an electron moving with a speed, v= 8.8 x105 m/s

charge of electron,which is moving positive direction,q=1.6x10-19 C

then the magnetic force is, F=q*v*B*sin(theta)

but all are mutually perpendicular direction. So, theta=90 then sin(90)=1

so that, rearrange the terms,

B=F/q*v

substitute the values in above,

B=(2.4 x10-13 N)/(1.6x10-19 C*8.8 x105 m/s)

=1.7 T

the magnitude of the magnetic field is positive direction.

the direction of the magnetic field is indicate z axis.