A magnetic field of 4.0×10−4k^T exerts a force of 4.0i^+3.0j^×10−10Non a particle having a charge of 10×10−9C and going in the X - Y plane. velocity of the particle is

4i^+3j^.10−10=10×10−9vxi^+vyj^×4×10−4k^ =10×10−9−vxj^+vyi^×4×10−4 =4×10−12−vxj^+vyi^ −vxj^+vyi^=4i^+3j^×10−104×10−12 =4i^+3j^×25 ⇒vx=75,vy=−100 ∴v¯=75i^−100j^m/s

A magnetic field of 4.0×10−4k^T exerts a force of 4.0i^+3.0j^×10−10Non a particle having a charge of 10×10−9C and going in the X - Y plane. velocity of the particle is

4i^+3j^.10−10=10×10−9vxi^+vyj^×4×10−4k^ =10×10−9−vxj^+vyi^×4×10−4 =4×10−12−vxj^+vyi^ −vxj^+vyi^=4i^+3j^×10−104×10−12 =4i^+3j^×25 ⇒vx=75,vy=−100 ∴v¯=75i^−100j^m/s

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A magnetic field of $(4.0 \times 10^{- 4} \hat{k}) T$ exerts a force of $(4.0 \hat{i} + 3.0 \hat{j}) \times 10^{- 10} N$ on a particle having a charge of $10 \times 10^{- 9} C$ and going in the X - Y plane. velocity of the particle is

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$(75 \hat{i} - 100 \hat{j}) m/s$

$Z e r o$

$(- 75 \hat{i} - 100 \hat{j}) m/s$

$(- 75 \hat{i} + 100 \hat{j}) m/s$

answer is B.

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Detailed Solution

$(4 \hat{i} + 3 \hat{j}) {.10}^{- 10} = 10 \times 10^{- 9} [(v_{x} \hat{i} + v_{y} \hat{j}) \times (4 \times 10^{- 4} \hat{k})] = 10 \times 10^{- 9} (- v_{x} \hat{j} + v_{y} \hat{i}) \times 4 \times 10^{- 4} = 4 \times 10^{- 12} (- v_{x} \hat{j} + v_{y} \hat{i}) - v_{x} \hat{j} + v_{y} \hat{i} = (4 \hat{i} + 3 \hat{j}) \times \frac{10^{- 10}}{4 \times 10^{- 12}} = (4 \hat{i} + 3 \hat{j}) \times 25 \Rightarrow v_{x} = 75, v_{y} = - 100 ∴ \bar{v} = (75 \hat{i} - 100 \hat{j}) m/s$