A particle of charge - 16 × {10^{ - 18}} coulomb moving with velocity 10,,m{s^{ - 1}} along x -axis

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4.Moving Charges and Magnetism

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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 magnitude ${10^4}\,\,V/m$ 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

${10^3}\,Wb/{m^2}$

C

${10^5}\,Wb/{m^2}$

D

${10^{16}}\,Wb/{m^2}$

(AIEEE-2003)

Solution

(b) Since particle is moving undeflected.
So $qE = qvB \Rightarrow B = E/v = \frac{{{{10}^4}}}{{10}} = {10^3}\,Wb/{m^2}$

Standard 12

Physics

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