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QJEEM13Ph9a88: JEE Main 2013 Physics - Class 12 JEEM13Ph9a88 Solution: A circular loop of radius lies center of the small loop is on…

A circular loop of radius 0.30.3 cmcm lies center of the small loop is on the axis of the bigger loop. The distance between their centers is 1515 cm.cm. If a current of 2.02.0 AA flows through the smaller loop, than the flux linked with bigger loop is

Multiple Choice Options

  • A 9.1×10−11 9.1 \times {10^{ - 11}}\, weber
  • B 6×10−11 6 \times {10^{ - 11}}\, weber
  • C 3.3×10−11 3.3 \times {10^{ - 11}}\, weber
  • D 6.6×10−9 6.6 \times {10^{ - 9}}\, weber

Step-by-Step Verified Solution

As we know, Magnetic flux, ϕ=B.A\phi = B.A

μ0(2)(20×10−2)22[(0.2)2+(0.15)2]×π(0.3×10−2)2{{{\mu _0}\left( 2 \right){{\left( {20 \times {{10}^{ - 2}}} \right)}^2}} \over {2\left[ {{{\left( {0.2} \right)}^2} + {{\left( {0.15} \right)}^2}} \right]}} \times \pi {\left( {0.3 \times {{10}^{ - 2}}} \right)^2}

On solving

=9.216×10−11=9.2×10−11  = 9.216 \times {10^{ - 11}} = 9.2 \times {10^{ - 11}}\,\, weber

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