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>> $A particle of charge 16 \times 10^-18C movin$

Question

A particle of charge $16 \times 1 0^{- 18} C$ moving with velocity $10 m / s$ 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 $10 V m^{- 1}$ is along the negative Z-axis. If the charged particle continues moving along the X-axis, the magnitude of $B$ is :

A

$1 W b / m^{2}$

B

$1 0^{5} W b / m^{2}$

C

$1 0^{6} W b / m^{2}$

D

$1 0^{- 3} W b / m^{2}$

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Updated on : 2022-09-05

Solution

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Correct option is A)

Since particle continues moving along X-axis, so there should be no force on the particle.
$F_{n e t} = 0$
$F_{e l e c t i c} - F_{m a g n e t i c} = 0$
$q E = q v B$
$E = v B$
$10 = 10 \times B$

$B = 1 W b - m^{2}$

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