Question

A proton enters a magnetic field of intensity $1.5\, Wb / m ^{2}$ with a velocity $2 \times 10^{7} m / s$ in a direction at an angle $30^{\circ}$ with the field. The force on the proton will be (charge on proton is $1.6 \times 10^{-19} C$ )

• A. $2.4 \times 10^{-12} N$
• B. $4.8 \times 10^{-12} N$
• C. $1.2 \times 10^{-12} N$
• D. $7.2 \times 10^{-12} N$

Answer 1

Answer: (A)

Here: $q=16 \times 10^{-19} C,\, B=15\, Wb / m ^{2}$,
$v=2 \times 10^{7} m / s,\, \theta=30^{\circ}$ or $\sin 30^{\circ}=\frac{1}{2}$
Force on proton is given by
$F =q v B \sin \theta$
$=1.6 \times 10^{-19} \times 2 \times 10^{7} \times 1.5 \times \frac{1}{2}$
$=2.4 \times 10^{-12} N$