Question

A particle having the same charge as of electron moves in a circular path of radius $0.5\text{cm}$ under the influence of a magnetic field of $0.5\text{T}$. If an electric field of $100\text{V/m}$ makes it to move in a straight path then the mass of the particle is
(Given charge of electron $=1.6\times {10}^{-19}\text{C}$)

• A. $1.6\times {10}^{-27}\text{kg}$
• B. $1.6\times {10}^{-19}\text{kg}$
• C. $2.0\times {10}^{-24}\text{kg}$
• D. $9.1\times {10}^{-31}\text{kg}$

Answer 1

The correct option is C $2.0\times {10}^{-24}\text{kg}$

As particle is moving along a circular path
$\therefore R=\frac{mv}{qB}$
$m=\frac{qBR}{v}...\left(i\right)$
When path is straight line, then
$F_E=F_B$

$qE=qvB$
$E=vB\Rightarrow v=\frac{E}{B}...\left(ii\right)$
From equation $\left(i\right)$ and $\left(ii\right)$

$m=\frac{q{B}^{2}R}{E}=\frac{1.6\times {10}^{-19}\times (0.5{)}^2\times 0.5\times {10}^{-2}}{100}$
$\therefore m=2.0\times {10}^{-24}\text{kg}$