Question

. An electron has a circular path of radius 0.01m in a perpendicular magnetic induction 10^–3 T find the speed of electron A) 1.76× 10^6 m/s B) 1.76 × 10^–6 m/s C) 1.76 m/s D) none

Answer 1

Force due to magnetic field = Bqv
Centripetal Force = mv²/r

Centripetal Force is provided by magnetic field.
So,
Force due to magnetic field = Centripetal Force
Bqv = mv² / r
v = Bqr / m
= (10^-3 T × 1.6 × 10^-19 C × 0.01 m) / (9.1 × 10^-31 kg)
= 1.76 × 10^6 m/s

Speed of electron is 1.76 × 10^6 m/s

(A) is the correct option.

Answer 2

Answer:

1.76*10^6

Explanation:

According to the Lorentz force law,

F = q(v*B) -- eqn 1

where q is the charge on the particle, v is the particle's velocity, and B is the magnetic field. Since we are given that v and B are perpendicular, we need not worry about the cross product, and this becomes

F = qvB

Now, we know F = ma, and for uniform circular motion, a=

r

v

2

. So we get, F =

r

mv

2

. -- eqn 2

Equating eqn 1 and eqn 2, we get,

v = rqB/m

The mass of an electron is 9.109×10

−31

kg, and its charge is -1.602×10

−31

C.

Substituting the values we get,

v =

9.109×10

−31

kg

0.01m×1.602×10

−31

C×10

−3

T

=1.76×10

6

m/s.

Hence, the speed of the electron is nearly 1.76×10

6

m/s.