Question

A particle of charge q, mass m, and speed v enters a uniform magnetic field of magnitude B that is directed down the page as shown.
If the particle enters the field at a 30-degree angle as shown, what is the radius of the circular path the particle makes?

• A. $\frac{\sqrt{3}mv}{2Bq}$
• B. $\frac{qBr}{2m}$
• C. $\frac{qBr}{m}$
• D. $\frac{2qBr}{m}$
• E. $\frac{\sqrt{3}qBr}{2m}$

Answer 1

Answer: (A)

$\frac{\sqrt{3}mv}{2Bq}$

From figure, $v_{\perp }=v$ $cos{30}^o=\frac{\sqrt{3}v}{2}$

Radius of circular path $r=\frac{m{v}_{\perp }}{Bq}$

$⟹$ $r=\frac{\sqrt{3}mv}{2Bq}$