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