Question

A particle of charge $−16 × 10^{−18}$ coulomb moving with velocity $10\, ms^{−1}$ 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 $104 \,V/m$ 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 $E_{y}=v_{x}B_{z}$
$\Rightarrow B_{z}=\frac{E_{y}}{v_{z}}=\frac{10^{4}}{10}=10^{3}\,Wb /m^{2}$