Question

A proton enters a perpendicular magnetic field of 20 Tesla. If the velocity of proton is $4\times {10}^{7}m/s$ then the force acting on it will be :

• A. 1.28 Newton
• B. $1.28\times {10}^{-11}$ Newton
• C. 12.8 Newton
• D. $12.8\times {10}^{-11}$ Newton

Answer 1

Answer: (D)

$12.8\times {10}^{-11}$ Newton

$F=qVB$

so we have $B=20Tesla$ and $V=4\times {10}^7$

So putting this value

$Force=(1.6\times {10}^{-19})\left(20\right)(4\times {10}^7)$

$Force=12.8\times {10}^{-11}$ N