# A particle of mass m = 1.6 x 10-27 kg and charge q = 1.6 x 10-19 C enters a region of uniform magnetic field of strength 1  tesla  along the direction shown in fig. The speed of the particle is m/s . (i) The magnetic field is directed along the inward normal to the plane of the paper. The particle leaves the region of the field at the point F. Find the distance EF and the angle θ (ii) If the direction of the field is along the outward normal to the plane of the paper, find the time spent by the particle in the region of the magnetic field after entering it at E

Navjyot Kalra
8 years ago
Hello Student,
m = 1. 6 x 10-27 kg, q = 1.6 x 10-19 C
B = 1 T
v = 107 m/s
F = q.v B sin α
(acting towards O by Fleming’s left hand rule)
⇒ F = qv B [∵ α = 90°]
But F = ma
∴ qv B = ma ∴ a = qvB / m
= 1 . 6 x 10-19 x 107 x 1 / 1. 6 x 10-27 = 1.15 m/s2
∠ OEF = 45° (∵ OE act as a radius)
By symmetry ∠ EOF = 45°
∴ ∠EOF = 90° (by Geometry)
This is the centripetal acceleration
∴ v2 / r = 1015 ⇒ r = 1014 / 1015 = 0.1 m
Therefore EF = 0.141 m.
If the magnetic field is in the outward direction and the article enters in the same way at E, then according to Fleming’s left hand rule, the particle will turn towards clockwise direction and cover 3 /4 the of a circle as shown in the figure.
∴ The required = 3 / 4 x [ 2 πr / v] = 4.71 x 10-8 sec.
Thanks
Navjot Kalra