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4.Moving Charges and Magnetism
medium
A particle of charge $16\times10^{-16}\, C$ moving with velocity $10\, ms^{-1}$ along $x-$ axis enters a region where magnetic field of induction $\vec B$ is along the $y-$ axis and an electric field of magnitude $10^4\, Vm^{-1}$ is along the negative $z-$ axis. If the charged particle continues moving along $x-$ axis, the magnitude of $\vec B$ is
A
$16\times10^3\, Wb \,m^{ -2}$
B
$2\times10^3\, Wb \,m^{ -2}$
C
$1\times10^3\, Wb \,m^{ -2}$
D
$4\times10^3\, Wb \,m^{ -2}$
(JEE MAIN-2013)
Solution
Since particle is moving undeflected So, $qE\, = qvB$
$ \Rightarrow B = \frac{E}{V} = \frac{{{{10}^4}}}{{10}} = {10^3}\,wb/{m^2}$
Standard 12
Physics
