A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm . The centre of the small loop is on the axis of the big - Physics MCQ JEE Main 2026 | ExamTest.live

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A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm . The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm . If a current of 2.0 A flows through the smaller loop, then the flux linked with bigger loop is:

Solution & Step-by-step Explanation

Use the property of mutual induction: Φ_{2} = M I_{1} . M is same whether current is in small or big loop.Assume current I_{2} in the big loop. Magnetic field at axis at distance x is: B = \frac{μ _{0} I _{2} R ^{2}}{2 ( R ^{2} + x ^{2} ) ^{3/2}} Flux through small loop Φ_{s ma ll} = B \cdot Area_{s ma ll} = [\frac{μ _{0} I _{2} R ^{2}}{2 ( R ^{2} + x ^{2} ) ^{3/2}}] π r^{2} . M = \frac{Φ _{s ma ll}}{I _{2}} = \frac{μ _{0} π R ^{2} r ^{2}}{2 ( R ^{2} + x ^{2} ) ^{3/2}} .Substitute values: R = 0.2 m, r = 0.003 m, x = 0.15 m . Φ = M I = \frac{( 4 π \times 1 0 ^{- 7} ) π ( 0.2 ) ^{2} ( 0.003 ) ^{2} ( 2.0 )}{2 (( 0.2 ) ^{2} + ( 0.15 ) ^{2} ) ^{3/2}} Denominator = 2 (0.04 + 0.0225)^{3/2} = 2 (0.0625)^{3/2} = 2 (0.25)^{3} = 2 (0.015625) = 0.03125 .Numerator \approx 4 \times 1 0^{- 7} \times 9.87 \times 0.04 \times 9 \times 1 0^{- 6} \times 2 \approx 2.84 \times 1 0^{- 12} . Φ \approx 9.1 \times 1 0^{- 11} Wb .

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A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm . The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm . If a current of 2.0 A flows through the smaller loop, then the flux linked with bigger loop is:

6 \times 10^{-11},\text{Wb}

3.3 \times 10^{-11},\text{Wb}

6.6 \times 10^{-9},\text{Wb}

9.1 \times 10^{-11},\text{Wb}

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