JEE Main & Advanced Sample Paper JEE Main Sample Paper-12

  • question_answer
    In given figure, a wire loop has been bent so that it has three segments: segment ab (a quarter circle), be (a square comer), and a ca (straight). Here are three choices for a magnetic field through the loop                        
    (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) \[{{\vec{B}}_{3}}=2\hat{i}-5t\hat{j}-12\hat{k}\]
    where \[\vec{B}\] is in milliteslas and t is in seconds. If the induced current in the loop due to \[{{\vec{B}}_{1}},{{\vec{B}}_{2}}\] and \[{{\vec{B}}_{3}}\] are \[{{i}_{1}},{{i}_{2}}\] and \[{{i}_{3}}\] respectively, then

    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\[{{B}_{2}}>{{B}_{1}}>{{B}_{3}}\]


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