Question

A proton enters a magnetic field of flux density $1.5 \,Wb / m ^{2}$ with a speed of $2 \times 10^{7} m / s$ at angle of $30^{\circ}$ with the field. The force on a proton will be

• A. $0.24 \times 10^{-12} N$
• B. $2.4 \times 10^{-12} N$
• C. $24 \times 10^{-12} N$
• D. $0.024 \times 10^{-12} N$

Answer 1

Answer: (B)

Magnetic force,
$F=q v\, B\, \sin\, \theta$
$\therefore F=\left(1.6 \times 10^{-19}\right) \times\left(2 \times 10^{7}\right) \times(1.5) \sin 30^{\circ}$
$F=1.6 \times 10^{-12} \times 2 \times 1.5 \times \frac{1}{2}$
$F=2.4 \times 10^{-12} N$