Question

A proton moving with a velocity, $2.5 \times 10^{7} m / \sec$. enters a magnetic field of intensity $2.5 T$ making an angle $30^{\circ}$ with the magnetic field. The force on the proton is

• A. $3 \times 10^{-12} N$
• B. $5 \times 10^{-12} N$
• C. $6 \times 10^{-12} N$
• D. $9 \times 10^{-12} N$

Answer 1

Answer: (B)

Velocity of proton, $v=25 \times 10^{7} m / s$
Magnetic field, $B=2.5\, T$
$\theta=30^{\circ}$
Magnetic force on proton in magnetic field is given as
$F =B q v \sin \theta$
$=2.5 \times 1.6 \times 10^{-19} \times 25 \times 10^{7} \sin 30^{\circ}$
$=6.25 \times 1.6 \times 10^{-12} \times \frac{1}{2}$
$=5 \times 10^{-12} N$