Question

A particle of charge $-16\times {10}^{-18}$ coulomb moving with velocity 10 $m{s}^{-1}$ 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 ${10}^{4}V/m$ is along the negative $z$-axis. lf 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

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

Particle travels along $x$ axis

Hence $V_y=V_z=0$

Electric feild is along the negative $z$ axis

$E_x=E_y=0$

Therefore net force on the particle $\overrightarrow{F}=q(\overrightarrow{E}+V\times \overrightarrow{B})$

Resolve the motion along three coordinate axis

$a_x=\frac{F_x}{m}=\frac{q}{m}(E_x+V_y{B}_z-V_x{B}_y)a_y=\frac{F_x}{m}=\frac{q}{m}(E_y+V_z{B}_x-V_{xz})a_z=\frac{F_x}{m}=\frac{q}{m}(E_z+V_x{B}_y-V_y{B}_x)$

Since,

$V_y=V_z=0E_x=E_y=0B_x=B_z=0$

$a_x=a_y=0a_z=\frac{q}{m}(-E_z+V_x{B}_y)$

Again $a_z=0$ , as the particle transverse through the region undeflected

$E_z=V_x{B}_y$

$B_y=\frac{E_z}{V_x}=\frac{{10}^4}{10}={10}^{3}wb/m^2$