(1) \[{{\vec{B}}_{1}}=3\hat{i}+7\hat{j}-5t\hat{k}\] |
(2) \[{{\vec{B}}_{2}}=5t\hat{i}-4\hat{j}-15\hat{k}\] |
(3) \[{{\vec{B}}_{3}}=2\hat{i}-5t\hat{j}-12\hat{k}\] |
A) \[{{i}_{1}}>{{i}_{2}}>{{i}_{3}}\]
B) \[{{i}_{2}}>{{i}_{1}}>{{i}_{3}}\]
C) \[{{i}_{3}}>{{i}_{2}}>{{i}_{1}}\]
D) \[{{i}_{1}}={{i}_{2}}={{i}_{3}}\]
Correct Answer: B
Solution :
Here, \[\text{are}{{\text{a}}_{\text{square}}}\text{are}{{\text{a}}_{\text{quarter}}}_{\text{circle}}\text{are}{{\text{a}}_{\text{triangle}}}\] so \[\frac{d\phi }{dt}\]of\[{{B}_{2}}>{{B}_{1}}>{{B}_{3}}\]You need to login to perform this action.
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