Question

An electron has a circular path of radius 0.01 m in a perpendicular magnetic induction ${10}^{-3}$ T. The speed of the electron is nearly :

• A. $1.76\times {10}^4$ m\s
• B. $1.76\times {10}^6$ m\s
• C. $3.52\times {10}^8$ m\s
• D. $7.04\times {10}^6$ m\s

Answer 1

The correct option is B $1.76\times {10}^6$ m\s

Answer is B.
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, $a=\frac{v^2}{r}.$ So we get, F = $\frac{{mv}^2}{r}$. -- eqn 2
Equating eqn 1 and eqn 2, we get,
v = rqB/m
The mass of an electron is $9.109\times {10}^{-31}kg$, and its charge is -$1.602\times {10}^{-31}C$.
Substituting the values we get,
v = $\frac{0.01m\times 1.602\times {10}^{-31}C\times {10}^{-3}T}{9.109\times {10}^{-31}kg}=1.76\times {10}^{6}m/s$.
Hence, the speed of the electron is nearly $1.76\times {10}^{6}m/s$.