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. \[{{\vec{F}}_{e}}+{{\vec{F}}_{m}}=0\] \[\therefore \] \[16\times {{10}^{-14}}=16\times {{10}^{-17}}B\] \[\Rightarrow \] \[B=\frac{16\times {{10}^{-14}}}{16\times {{10}^{-17}}}={{10}^{3}}\] \[B={{10}^{3}}\,Wb/{{m}^{2}}\]You need to login to perform this action.
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