Electromagnetic Induction

JEE Main level, numerical answer, medium

127 of 153

Question 127 of 153

Two concentric circular coils with radii 1cm1cm and 1000cm1000cm and number of turns 1010 and 200200 respectively are placed coaxially with centers coinciding. The mutual inductance of this arrangement will be _____ ×10–8H\times {10}^{–8}H. (Take, π2=10{\pi }^{2}=10)

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Answer: 4

Let the magnetic field of the 200200 turns be B1{B}_{1}. The formula of magnetic field is given by B=μ0i2rB=\dfrac{{\mu }_{0}i}{2r}
Let r=1000cmr=1000cm. Thus,
B1=nμ0×I2×10=200μ0I2×10{B}_{1}=\dfrac{n{\mu }_{0}\times I}{2\times 10}=\dfrac{200{\mu }_{0}I}{2\times 10}
The flux through the coil of radius r1=1cm{r}_{1}=1cm is ϕ=10×(B⃗⋅A⃗)\phi =10\times \left(\vec{B}\cdot \vec{A}\right)=10×B1πr21=10\times {B}_{1}\pi {{r}^{2}}_{1}
⇒ϕ=10×200μ0I2×10×π(0.01)2\Rightarrow \phi =10\times \dfrac{200{\mu }_{0}I}{2\times 10}\times \pi {\left(0.01\right)}^{2}
L=10×200μ02×10×π(0.01)2=10×200(4π×10−7)2×10×π(0.01)2=4×10−8HL=10\times \dfrac{200{\mu }_{0}}{2\times 10}\times \pi {\left(0.01\right)}^{2} =10\times \dfrac{200\left(4\pi \times {10}^{-7}\right)}{2\times 10}\times \pi {\left(0.01\right)}^{2} =4\times {10}^{-8}H

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