question_answer

A particle of charge \[-16\times {{10}^{-18}}C\] moving with velocity \[10m{{s}^{-1}}\] 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     AIEEE  Solved  Paper-2003

A) \[{{10}^{3}}\] Wb / \[{{m}^{2}}\]     

B)       \[{{10}^{5}}\] Wb/\[{{m}^{2}}\]                

C) \[{{10}^{16}}\] Wb/\[{{m}^{2}}\]      

D)       \[{{10}^{-3}}\] Wb/\[{{m}^{2}}\]

Correct Answer: A

Solution :

The Lorentz's force on the charge particle is             So,              F = q (E + v \[\times \] B) or                F = \[{{F}_{e}}+{{F}_{m}}\]             \[\therefore \]      \[{{F}_{e}}=qE=-16\times {{10}^{-18}}\times {{10}^{4}}(-\hat{k})\]                                 \[=16\times {{10}^{-14}}\hat{k}\] and            \[_{m}=-16\times {{10}^{-18}}(10\,\hat{i}\times B\,\hat{j})\]                     \[=-16\times {{10}^{-17}}\times B\,(\hat{k})\]                     \[=-16\times {{10}^{-17}}B\,\hat{k}\] Since, particle will continue to move along + x-axis, so resultant force is equal to zero.                     \[{{F}_{e}}+{{F}_{m}}=0\] \[\therefore \]      \[16\times {{10}^{-14}}=16\times {{10}^{-17}}B\] \[\Rightarrow \]   \[B=\frac{16\times {{10}^{-14}}}{16\times {{10}^{-17}}}={{10}^{3}}\]                 \[B={{10}^{3}}Wb/{{m}^{2}}\]