Question

An electron whose $\frac{e}{m}$ is $1.76\times {10}^{11}\text{C/kg}$ enters a region of a uniform magnetic field of induction $2\times {10}^{-3}\text{T}$ with velocity of $3\times {10}^6{\text{ms}}^{-1}$ in a direction making an angle of ${45}^{\circ }$ with the field. The pitch of its helical path in the region is

• A. $8.4\text{cm}$
• B. $5.36\text{cm}$
• C. $3.8\text{cm}$
• D. $1.5\text{cm}$

Answer 1

The correct option is C $3.8\text{cm}$

Given:
$\theta ={45}^{\circ };v=3\times {10}^6{\text{ms}}^{-1}$
$\frac{e}{m}=1.76\times {10}^{11}\text{C/kg};B=2\times {10}^{-3}\text{T}$

Time period of motion of charged particle in magnetic field,

$T=\frac{2\pi m}{qB}=\frac{2\pi }{1.76\times {10}^{11}\times 2\times {10}^{-3}}$

And we know that pitch of the helix will be

$P=T\times v_{||}=T\times v\cos {45}^{\circ }$

$P=\frac{2\pi \times 3\times {10}^6\times \left(1/\sqrt{2}\right)}{1.76\times {10}^{11}\times 2\times {10}^{-3}}$

$P=3.78\times {10}^{-2}\text{m}=3.8\text{cm}$

Hence, option $\left(c\right)$ is right.