An electron (mass 9 10-31 kg and charge 1.6 10-19 C) moving with speed c/100 (c = speed of light) is injected into a magnetic field B of magnitude 9 10-4 T perpendicular to its direction of motion. We wish to apply a uniform electric field E together with the magnetic field so that the electron does not deflect from its path. Then (speed of light c = 3 108 m s-1)
- E is parallel to B and its magnitude is 27 102 V m-1
- E is parallel to B and its magnitude is 27 104 V m-1
- E is perpendicular to B and its magnitude is 27 104 V m-1
- E is perpendicular to B and its magnitude is 27 102 V m-1
Answer
(4) E is perpendicular to B and its magnitude is 27 102 V m-1
Solution
For no deflection the electric force must cancel the magnetic force: q E = -q( v B)
So E is along v B, i.e. perpendicular to both v and B.
v = c100 = 3 106 m s-1
E = vB = 3 106 9 10-4 = 2700 = 27 102 V m-1
Answer: (4)