Question

A electron moving with a velocity ${10}^8\text{m}/\text{s}$ enters a magnetic field at an angle of ${30}^{\circ }$ to the direction of the field. Calculate the value of magnetic field so that radius of the helical path followed by the particle will be $2\text{m}$ ? (Mass of electron $m=9.1\times {10}^{-31}\text{kg}$)

• A. $1.42\times {10}^{-4}\text{T}$
• B. $0.71\times {10}^{-4}\text{T}$
• C. $0.71\times {10}^{-3}\text{T}$
  
• D. $2.84\times {10}^{-4}\text{T}$

Answer 1

The correct option is A $1.42\times {10}^{-4}\text{T}$

Velocity component perpendicular to the magnetic field,

$v_{\perp }={10}^8\times \sin {30}^{\circ }=0.5\times {10}^8\text{m}/\text{s}$

Given radius $r=2\text{m}$, and we know, $r=\frac{mv}{qB}$

$\Rightarrow 2=\frac{9.1\times {10}^{-31}\times 0.5\times {10}^8}{1.6\times {10}^{-19}\times B}$

$\Rightarrow 2=\frac{2.84\times {10}^{-4}}{B}$

$\therefore B=1.42\times {10}^{-4}\text{T}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option (a) is the correct answer.

Why this question ? Hint : Velocity component perpendicular to magnetic field $\left(v_{\perp }\right)$ is responsible for the circular motion, and velocity component parallel the to field $\left(v_{\parallel }\right)$ is responsible for the pitch.