Question

A particle of charge $-16\times 10^{-18}$ coulomb moving with velocity $10m{s}^{-1}$ along the $x$-axis enters a region where a magnetic field of induction $B$ is along the $y$ -axis, and an electric field of induction $B$ is along the y-axis, and an electric field of magnitude $104V/m$ is along the negative z-axis. If the charged particle continues moving along the x-axis, the magnitude of B is

• A. $10^{3}Wb/m^2$
• B. $10^{5}Wb/m^2$
• C. $10^{16}Wb/m^2$
• D. $10^{-3}Wb/m^2$

Answer 1

Answer: (A)

$\vec{F}=q\left(\vec{E}+\vec{v}\times \vec{B}\right)...\left(1\right)$
The solution of this problem can be obtained by resolving the motion along the three coordinate axes namely
$a_x=\frac{F_x}{m}=\frac{q}{m}\left(E_x+v_y{B}_z-v_z{B}_y\right)$
$a_y=\frac{F_y}{m}=\frac{q}{m}\left(E_y+v_z{B}_x-v_x{B}_z\right)$
$a_z=\frac{F_z}{m}=\frac{q}{m}\left(E_z+v_x{B}_y-v_y{B}_z\right)$
For the given problem,
$E_x=E_y=0,v_y=v_z=0$ and $B_x=B_z=0$
Substituting in equation $\left(2\right)$, we get
$a_x=a_z=0$ and $a_y=E_y-v_x{B}_z$
If the particle passes through the region undeflected ay is also zero, then $E_y=v_x{B}_z$
$\Rightarrow B_z=\frac{E_y}{v_z}=\frac{10^4}{10}=10^{3}Wb/m^2$