Question

A particle of charge $-16\times {10}^{-18}C$ moving with velocity $10m/s$ along the $X-$axis enters region where a magnetic field of induction $B$ is along $Y-$axis and field of magnitude ${10}^{4}V/m$ is along the negative $Z-$axis. If the charged particle continuous moving along the $X-$axis, the magnitude of $B$ is:

• A. ${10}^{6}Wb/m^2$
• B. ${10}^{5}Wb/m^2$
• C. ${10}^{3}Wb/m^2$
• D. ${10}^{7}Wb/m^2$

Answer 1

The correct option is C ${10}^{3}Wb/m^2$

Given that,

Particle of charge$q=1.6\times {10}^{-19}C$

Velocity $v=10m/s$

Field induction $B={10}^{4}V/m$

Now, particle travel along x-axis

Then, $v_y=v_z=0$

And field induction B is along y-axis

Then, $B_x=B_z=0$

Electric field is along negative z-axis

Then, $E_x=E_y=0$

Now, net force on particle is

$F=q(\overrightarrow{E}+\overrightarrow{v}\times \overrightarrow{B})$

Resolve the motion along the three coordinates

$F_x=m{a}_x$

$a_x=\frac{F_x}{m}=\frac{q}{m}(E_x+v_y{B}_z-v_z{B}_y)$

$a_y=\frac{q}{m}(E_y+v_z{B}_x-v_x{B}_z)$

$a_z=\frac{q}{m}(E_z+v_x{B}_y-v_x{B}_y)$

So, $E_x=E_y=0,v_y=v_z=0,B_x=B_z=0$

Now,

$a_x=a_y=0$

$a_z=\frac{q}{m}(-E_z+v_x{B}_y)$

Now, again

$a_z=0$ the particle traverse through the region un-deflected

$E_z=v_x{B}_y$

$B_y=\frac{E_z}{v_x}$

$B_y=\frac{{10}^4}{10}$

$B_y={10}^{3}wb/m^2$

Hence, the magnitude of B is ${10}^{3}wb/m^2$