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JEE Main 20236 AprilEvening Shift
Exam
JEE Main
Year
2023
Date
6 April
Shift
Evening Shift
Subject
Physics
Chapter
Electromagnetic Induction

JEE Main 2023 · 6 April · Evening Shift · Physics · Electromagnetic Induction

Two concentric circular coils with radii 1cm and 1000cm and number...

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)

Correct answer

0

Step-by-step solution

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×(B→⋅A→)ϕ = 10 \times \left(\overset{\to}{B} \cdot \overset{\to}{A}\right)= 10 \times {B}_{1}\pi{r}^{2}1</p><p></p><p>⇒ ϕ = 10 \times \dfrac{200{\mu}_{0}I}{2 \times 10} \times \pi{\left(0 . 01\right)}^{2}</p><p></p><p>L = 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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