JEE Main Electromagnetic Induction PYQ

1. (JEE Main 2024 (Online) 5th April Morning Shift)

Two conducting circular loops A and B are placed in the same plane with their centres coinciding as shown in figure. The mutual inductance between them is:

JEE Main 2024 (Online) 5th April Morning Shift Physics - Electromagnetic Induction Question 3 English

A. $\frac{μ_{o}}{2 π} \cdot \frac{b^{2}}{a}$

B. $\frac{μ_{o} π a^{2}}{2 b}$

C. $\frac{μ_{0} π b^{2}}{2 a}$

D. $\frac{μ_{0}}{2 π} \cdot \frac{a^{2}}{b}$

Correct Option is (B)

$\begin{aligned} \frac{ϕ_{A / B}}{I_{B}} = M \\ \Rightarrow M = \frac{μ_{0} I_{B} \times π a^{2}}{2 b \times I_{B}} \\ = \frac{μ_{0} π a^{2}}{2 b} \end{aligned}$

2. (JEE Main 2024 (Online) 31st January Morning Shift)

A small square loop of wire of side $l$ is placed inside a large square loop of wire of side $L (L = l^{2})$ . The loops are coplanar and their centers coincide. The value of the mutual inductance of the system is $\sqrt{x} \times 10^{- 7} H$ , where $x =$ _________.

Correct answer is 128

JEE Main 2024 (Online) 31st January Morning Shift Physics - Electromagnetic Induction Question 10 English Explanation

Flux linkage for inner loop.

$\begin{aligned} ϕ = B_{center} \cdot ℓ^{2} \\ = 4 \times \frac{μ_{0} i}{4 π \frac{L}{2}} (\sin 45 + \sin 45) ℓ^{2} \\ ϕ = 2 \sqrt{2} \frac{μ_{0} i}{π L} ℓ^{2} \\ M = \frac{ϕ}{i} = \frac{2 \sqrt{2} μ_{0} ℓ^{2}}{π L} = 2 \sqrt{2} \frac{μ_{0}}{π} \\ = 2 \sqrt{2} \frac{4 π}{π} \times 10^{- 7} \\ = 8 \sqrt{2} \times 10^{- 7} H \\ = \sqrt{128} \times 10^{- 7} H \\ x = 128 \end{aligned}$

3. (JEE Main 2024 (Online) 27th January Morning Shift)

Two coils have mutual inductance $0.002 H$ . The current changes in the first coil according to the relation $i = i_{0} \sin ω t$ , where $i_{0} = 5 A$ and $ω = 50 π$ rad/s. The maximum value of emf in the second coil is $\frac{π}{α} V$ . The value of $α$ is _______.

Correct answer is 2

$\begin{aligned} ϕ = Mi = {Mi}_{0} \sin ω t \\ EMF = - M \frac{di}{dt} = - 0.002 (i_{0} ω \cos ω t) \\ {EMF}_{max} = i_{0} ω (0.002) = (5) (50 π) (0.002) \\ {EMF}_{max} = \frac{π}{2} V \end{aligned}$

4. (JEE Main 2023 (Online) 29th January Morning Shift)

Find the mutual inductance in the arrangement, when a small circular loop of wire of radius ' $R$ ' is placed inside a large square loop of wire of side $L$ $(L ≫ R)$ . The loops are coplanar and their centres coincide :

JEE Main 2023 (Online) 29th January Morning Shift Physics - Electromagnetic Induction Question 36 English

A. $M = \frac{\sqrt{2} μ_{0} R}{L^{2}}$

B. $M = \frac{2 \sqrt{2} μ_{0} R}{L^{2}}$

C. $M = \frac{2 \sqrt{2} μ_{0} R^{2}}{L}$

D. $M = \frac{\sqrt{2} μ_{0} R^{2}}{L}$

Correct Option is (C)

JEE Main 2023 (Online) 29th January Morning Shift Physics - Electromagnetic Induction Question 36 English Explanation
$\begin{aligned} B at centre = \frac{μ_{0} i}{4 π (\frac{L}{2})} (\frac{2}{\sqrt{2}}) \times 4 \\ = \frac{\sqrt{2} μ_{0} i}{2 π L} \times 4 \\ = (\frac{2 \sqrt{2} μ_{0} i}{π L}) \end{aligned}$

Mutual inductance $= \frac{B \cdot A}{i}$

$= \frac{2 \sqrt{2} μ_{0} i}{π L} \times \frac{π R^{2}}{i}$

$= (\frac{2 \sqrt{2} μ_{0} R^{2}}{L})$

5. (JEE Main 2023 (Online) 6th April Evening Shift)

Two concentric circular coils with radii $1 cm$ and $1000 cm$ , and number of turns 10 and 200 respectively are placed coaxially with centers coinciding. The mutual inductance of this arrangement will be ___________ $\times 10^{- 8} H$ . (Take, $π^{2} = 10$ )

Correct answer is 4

The magnetic field $B_{2}$ due to the current $I_{2}$ in the larger coil with 200 turns is given by:

$B_{2} = \frac{N_{2} μ_{0} I_{2}}{2 r_{2}} = \frac{200 μ_{0} I_{2}}{2 \times 10}$

The magnetic flux $ϕ_{1, 2}$ through the smaller coil due to this magnetic field is given by:

$ϕ_{1, 2} = N_{1} {\vec{B}}_{2} \cdot {\vec{A}}_{1} = N_{1} N_{2} \frac{μ_{0} I_{2}}{2 r_{2}} \cdot π r_{1}^{2}$

Since $ϕ_{1, 2} = M I_{2}$ , we can solve for the mutual inductance $M$ :

$M = \frac{N_{1} N_{2} \frac{μ_{0} I_{2}}{2 r_{2}} \cdot π r_{1}^{2}}{I_{2}}$

Substituting the given values for $r_{1}$ , $N_{1}$ , $r_{2}$ , and $N_{2}$ :

$M = \frac{10 \times 200 \times 4 π \times 10^{- 7} \times π \times (0.01)^{2}}{2 \times 10}$

Simplifying the expression, we get:

$M = 4 \times 10^{- 8} H$

So, the mutual inductance between the two concentric coils is $4 \times 10^{- 8} H$ .