Question

A proton enters a magnetic field with a velocity of $2.5\times {10}^{7}m{s}^{-1}$ making an angle ${30}^0$ with the magnetic field. The force on the proton is $(B=25T)$

• A. $1.25\times {10}^{-11}N$
• B. $2.5\times {10}^{-11}N$
• C. $5.0\times {10}^{-11}N$
• D. $7.5\times {10}^{-11}N$

Answer 1

Answer: (C)

$5.0\times {10}^{-11}N$

$\overline{F}=q(\overrightarrow{v}\times \overrightarrow{B})$
$F-qvBsin\theta$
$F=1\cdot 6\times {10}^{-19}\times 2\cdot 5\times {10}^7\times 25\times sin30$
$F=5\times {10}^{-11}N$