NEET AIPMT SOLVED PAPER SCREENING 2009

  • question_answer
    A conducting circular loop is placed in a uniform magnetic field 0.04 T with its plane perpendicular to the magnetic field. The radius of the loop starts shrinking at 2 \[=\frac{\sigma }{{{\varepsilon }_{0}}}\left( \frac{{{a}^{2}}}{c}-\frac{{{b}^{2}}}{c}+c \right)=\frac{\sigma }{{{\varepsilon }_{0}}}(2a)\,(\because c=a+b)\]. The induced emf in the loop when the radius is 2 cm is

    A) \[{{V}_{A}}={{V}_{C}}\ne {{V}_{B}}\]

    B)                        \[\therefore \]

    C) \[{{v}_{S}}=v+{{v}_{B}}\]                             

    D) \[=\frac{1000}{100}=10\,\text{m}{{\text{s}}^{-1}}\]

    Correct Answer: A

    Solution :

    Key Idea According to Faraday's second law of electromagnetic induction the induced emf is given by rate of change of magnetic flux linked with the circuit. Here, B = 0.04 T and \[C{{H}_{3}}Cl\] Induced emf, \[AlC{{l}_{3}}\] \[n=4,l=0,m=0,s=-1/2\] Now, if r = 2 cm \[n=5,l=3,m=0,s=+1/2\] \[n=3,l=2,m=-3,s=-1/2\]


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