Question

An electron moves with a velocity 1 × 103m/* in a magnetic field of induction 0.3 T at an angle 30°. If e/m of electron is 1.76 × 1011 C/kg, the radius of the path is nearly :

(A) 10–9 m

(B) 2 × 10–8 m

(C) 10–6 m

(D) 10–10 m

Answer 1

Correct Answer is: (B) 2 × 10–8 m

If electron moves in a magnetic field at an angle θ (other than 0°, 180° or 90°), its velocity can be resolved in two components one along Vector B and another perpendicular to Vector B. Let the two components be V|and v⊥ . Then

The component perpendicular to field ( v⊥ )gives a circular path and the component parallel to field (V||) gives a straight line path. The resultant path is, helix as shown in figure.

The radius of this helical path is

Given, v = 1 × 103 m/*, B = 0.3T, θ = 30°

e/m = 1.76 × 1011 C/kg