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  • A particle of mass 'm' and charge 'q' has an initial velocity    If an electric field&nb

A particle of mass 'm' and charge 'q' has an initial velocity \vec{v}=v_{0} \hat{j}   If an electric field \overrightarrow{\mathrm{E}}=\mathrm{E}_{0} \hat{i}  and magnetic field \overrightarrow{\mathrm{B}}=\mathrm{B}_{0} \hat{i}  act on the particle, its speed will double after a time: 
Option: 1 \frac{\sqrt{3}mv_{0}}{qE_{0}}
 Option 2 \frac{{3}mv_{0}}{qE_{0}}
Option:3 \frac{\sqrt{2}mv_{0}}{qE_{0}}  
Option: 4\frac{{2}mv_{0}}{qE_{0}}

Answers (1)

best_answer

 

Initially \overrightarrow{\mathrm{v}}=\mathrm{v}_{0} \hat{\mathrm{j}}

As both electric and magnetic field is in the x-direction

And initial velocity is perpendicular to both electric and magnetic fields.

So particle will move in a helical motion

and magnitude of velocity does not change in the y–z plane

and Speed will be increased due to electric field only i.e in the x-direction

and finally, the speed of the particle becomes double the initial speed.

i.e \left(2 \mathrm{v}_{0}\right)^{2}=\mathrm{v}_{0}^{2}+\mathrm{v}_{\mathrm{x}}^{2} \quad ; \quad \mathrm{v}_{\mathrm{x}}=\sqrt{3} \mathrm{v}_{0}

And in x-direction F_x=qE\Rightarrow a_x=\frac{qE}{m}

So v_x=\sqrt{3} \mathrm{v}_{0}=0+\frac{\mathrm{qE}}{\mathrm{m}} \mathrm{t} ; \quad \mathrm{t}=\frac{\mathrm{mv}_{0} \sqrt{3}}{\mathrm{qE}}

Hence the correct option is (1). 

Posted by

Ritika Jonwal

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