Question

An electron is moving perpendicular to a magnetic field of strength 10 tesla with a velocity of 2200 $kmse{c}^{-1}$. Calculate the radius of its path. (The mass of the electron $m=9.1\times {10}^{-31}kg$ and its charge $e$ = 1.6 $\times {10}^{-19}C$ )

• A. 2.50 $\times {10}^{-3}m$
• B. 1.42 $\times {10}^{-2}m$
• C. 1.25 $\times {10}^{-6}m$
• D. 1.78 $\times {10}^{-3}m$

Answer 1

The correct option is C 1.25 $\times {10}^{-6}m$

Given:
Magnetic field, $B=10T$;
velocity of the electron, $v=2200km{s}^{-1}$ = $22\times {10}^{5}m{s}^{-1}$ and
magnitude of the charge of an electron, $e$ = 1.6 $\times {10}^{-19}$C.
The force F is equal to the centripetal force acting on the electron.
$⟹$ F = $e$vB = $\frac{m_e{v}^2}{r}$ --- (1);
where $r$ is the radius of the path of the electron and $m_e$ is the mass of the electron = 9.1 $\times {10}^{-31}kg$.
On solving equation (1), $r=\frac{m_{e}v}{e\text{B}}$
Substituting the values given, we get: $r=\frac{9.1\times {10}^{-31}kg\times 22\times {10}^{5}m{s}^{-1}}{1.6\times {10}^{-19}C\times 10T}$
$=1.25\times {10}^{-6}m$.