Question

A positive charge moves with constant velocity to the right through both a magnetic and electrostatic field. Which of the following conditions would allow for this to happen? Ignore effects from the force of gravity.

• A. B-field toward bottom of screen E-field toward bottom of screen.
• B. B-field to the right side of the screen E-field toward the bottom of the screen.
• C. B-field out of the screen E-field toward the top of the screen.
• D. B-field out of the screen E-field toward the bottom of the screen.

Answer 1

The correct option is C B-field out of the screen E-field toward the top of the screen.

The force on a charge q of mass m moving with a velocity v in a magnetic field perpendicularly is given by ,

$F_B=qvB$ ,

electric force on charge q by a field is given by ,

$F_E=qE$ ,

as the velocity of charge is constant i.e.net force on charge q is zero therefore both the fields should be equal and opposite ,

$F_B=F_E$ ,

now if we check all the given options , we find that in option B the magnetic and electric forces are opposite to each other , when B-field is right of the screen then by Fleming's left hand rule (taking motion of +ive particle as direction of current) we find magnetic force in upward direction , and to cancel this the E-field must be toward the bottom of the screen so that electric force on +ive charge is downward , toward the bottom . The next condition is B-field out of the screen , E-field toward the top of screen .