Question

A proton enters a region of constant magnetic field, $B$, of magnitude $1.0$ tesla with initial speed of $1.5\times {10}^{6}m/s$ . If the protons initial velocity vector makes an angle of ${30}^o$ with the direction of $B$. Calculate the speed of the protons $4$ seconds after entering the magnetic field.

• A. $5.0\times {10}^{5}m/s$
• B. $7.5\times {10}^{5}m/s$
• C. $1.5\times {10}^{6}m/s$
• D. $3.0\times {10}^{6}m/s$
• E. $6.0\times {10}^{6}m/s$

Answer 1

Answer: (C)

$1.5\times {10}^{6}m/s$

The magnetic force does not change the speed of a charged particle, it only changes the velocity i.e. the direction of charged particle ,

The magnetic force on a charge $q$ is given by $\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$

The acceleration by this force on of mass m of proton will be$\overrightarrow{a}=q(\overrightarrow{v}\times \overrightarrow{B})/m$

The speed will change only if any component of $\overrightarrow{a}$ is in the direction of velocity but cross product shows that acceleration is perpendicular to velocity so can only change the direction of velocity and speed will remain same as $1.5\times {10}^{6}m/s$