A) \[{{10}^{16}}\text{Wb/}{{\text{m}}^{\text{2}}}\]
B) \[{{10}^{5}}\text{Wb/}{{\text{m}}^{\text{2}}}\]
C) \[{{10}^{3}}\text{Wb/}{{\text{m}}^{\text{2}}}\]
D) \[{{10}^{-3}}\text{Wb/}{{\text{m}}^{\text{2}}}\]
Correct Answer: C
Solution :
As particle is moving without deviation, therefore \[Eq=Bqv\] \[\therefore \] \[B=\frac{E}{v}=\frac{{{10}^{4}}}{10}\] \[={{10}^{3}}Wb/{{m}^{2}}\]
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