A) \[16\times {{10}^{3}}\operatorname{Wb}{{\operatorname{m}}^{-2}}\]
B) \[2\times {{10}^{3}}\operatorname{Wb}{{\operatorname{m}}^{-2}}\]
C) \[1\times {{10}^{3}}\operatorname{Wb}{{\operatorname{m}}^{-2}}\]
D) \[4\times {{10}^{3}}\operatorname{Wb}{{\operatorname{m}}^{-2}}\]
Correct Answer: C
Solution :
[c] Since particle is moving un deflected |
So, \[{{q}_{E}}={{q}_{v}}B\] |
\[\Rightarrow \]\[B=\frac{E}{V}=\frac{{{10}^{4}}}{10}={{10}^{3}}wb/{{m}^{2}}\] |
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