question_answer

A particle of charge \[-16\times {{10}^{-18}}C\]moving with velocity \[10\,\text{m}{{\text{s}}^{-1}}\] 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 \[{{10}^{4}}\,\text{V}{{\text{m}}^{-1}}\] is along the negative\[z-\]axis. If the charged particle continues moving along the \[x-\]axis, then magnitude of B is :

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

B)  \[{{10}^{3}}Wb{{m}^{-2}}\]

C)  \[{{10}^{5}}Wb{{m}^{-2}}\]                       

D)  \[{{10}^{16}}Wb{{m}^{-2}}\]

Correct Answer: B

Solution :

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