Question

An electron having momentum $2.4 \times 10^{-23} \,kg \,ms ^{-1}$ enters a region of uniform magnetic field of $0.15 \,T$. The field vector makes an angle of $30^{\circ}$ with the initial velocity vector of the electron. The radius of the helical path of the electron in the field shall be

• A. $2\, mm$
• B. $1 \, mm$
• C. $\frac{\sqrt{3}}{2} \, mm$
• D. $0.5 \, mm$

Answer 1

Answer: (D)

The radius of the helical path of the electron in the uniform magnetic field is
$r=\frac{m v_{\perp}}{e B}$
$=\frac{m v \sin \theta}{e B}$
$=\frac{\left(2.4 \times 10^{-23}\, kg \,m s ^{-1}\right) \times \sin 30^{\circ}}{\left(1.6 \times 10^{-19} C \right) \times(0.15 T )}$
$=5 \times 10^{-4} m$
$=0.5 \times 10^{-3} m$
$=0.5\, mm$