Question

A proton is fired with a speed of $2\times {10}^{6}m/s$ at an angle of ${60}^{\circ }$ to the $X$ -axis. If a uniform magnetic field of $0.1$ tesla is applied along the $Y$-axis, the force acting on the proton is

• A. $1.603\times {10}^{-14}N$
• B. $1.6\times {10}^{-14}N$
• C. $3.203\times {10}^{-14}N$
• D. $3.2\times {10}^{-14}N$

Answer 1

Answer: (A)

$1.603\times {10}^{-14}N$

Given : A proton is fired with a speed of 2×106m/s2×106m/s at an angle of 60∘60∘ to the XX -axis. If a uniform magnetic field of 0.10.1 tesla is applied along the YY-axis, the force acting on the proton is

Charge on proton=$1.603\times {10}^{-19}$

Solution :

Formula for force on moving charge is $F=q(\overrightarrow{V}\times \overrightarrow{B})$

We have two components of velocity here one in Y-axis and other in X-axis . Magnetic field will act only on x component of velocity as Y component of velocity is parellal to Magnetic vector and cross product of parellal vector is 0.

Substituting all the values in the formula we get $F=(1.603\times {10}^{-19})\times (2\times {10}^{6}cos60)\times 0.1$

$F=1.603\times {10}^{-14}$

The Correct Opt=A