Question

A proton moving with a velocity, $2.5\times 10^{7}m/\sec$. enters a magnetic field of intensity $2.5T$ 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, $B=2.5T$
$\theta =30^{\circ }$
Magnetic force on proton in magnetic field is given as
$F=Bqv\sin \theta$
$=2.5\times 1.6\times 10^{-19}\times 25\times 10^7\sin 30^{\circ }$
$=6.25\times 1.6\times 10^{-12}\times \frac{1}{2}$
$=5\times 10^{-12}N$