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 :
[a] 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. |
\[{{F}_{e}}+{{F}_{m}}=0\] |
\[\therefore \] \[16\times {{10}^{-14}}=16\times {{10}^{-17}}B\] |
\[\Rightarrow \] \[B=\frac{16\times {{10}^{-14}}}{16\times {{10}^{-17}}}={{10}^{3}}\] |
\[B={{10}^{3}}Wb/{{m}^{2}}\] |
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