1. ⇒ (MHT CET 2023 13th May Morning Shift )

The magnet is moved towards the coil with speed ' $V$ '. The induced e.m.f. in the coil is ' $e$ '. The magnet and the coil move away from one another each moving with speed ' $V$ '. The induced e.m.f. in the coil is

A. e

B. 2e

C. $\frac{e}{2}$

D. $4 e$

Correct Option is (B)

The equation for the induced emf is:

$e = B / V$

Relative velocity between the coil and the magnet is:

$v_{r} = 2 v$

$∴$ The new induced emf in the coil is:

$e_{new} = B l \cdot 2 V = 2 e$

2. ⇒ (MHT CET 2023 12th May Evening Shift )

An air craft of wing span $40 m$ files horizontally in earth's magnetic field $5 \times 10^{- 5} T$ at a speed of $500 m / s$ . The e.m.f. generated between the tips of the wings of the air craft is

A. 0.5 V

B. 1V

C. 1.2V

D. 1.5V

Correct Option is (B)

The emf generated between the tips of the wings of the air craft is,

$\begin{aligned} ε = Blv \\ ε = 5 \times 10^{- 5} \times 40 \times 500 \\ ε = 1 V \end{aligned}$

3. ⇒ (MHT CET 2023 10th May Morning Shift )

A square loop of area $25 {cm}^{2}$ has a resistance of $10 Ω$ . This loop is placed in a uniform magnetic field of magnitude $40 T$ . The plane of loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in one second, will be

A. $1 \times 10^{- 4} J$

B. $1.0 \times 10^{- 3} J$

C. $5 \times 10^{- 3} J$

D. $2.5 \times 10^{- 3} J$

Correct Option is (B)

Given: area of square loop $= 25 cm$

$\begin{array}{ll} ∴ l = \sqrt{25} = 5 cm \Rightarrow 0.05 m \\ R = 10 Ω, t = 1 \sec, B = 40 T \\ ∴ Velocity v = \frac{l}{t} = \frac{0.05}{1} = 0.05 m / s \\ Motional emf ε_{max} = B / v \\ ∴ I = \frac{ε}{R} = \frac{B / v}{R} \\ ∴ \\ I = \frac{40 \times 0.05 \times 0.05}{10} = 0.01 A \end{array}$

We know, Force acting on loop

$\begin{aligned} ∴ F = BI l = 40 \times 0.01 \times 0.05 \\ = 0.02 N \\ Using W = F . s, \\ Work done W = BI l \times l \\ = 0.02 \times 0.05 \\ = 1 \times 10^{- 3} J \end{aligned}$

4. ⇒ (MHT CET 2023 9th May Evening Shift )

If current ' $I$ ' is flowing in the closed circuit with collective resistance ' $R$ ', the rate of production of heat energy in the loop as we pull it along with a constant speed ' $V$ ' is ( $L =$ length of conductor, $B =$ magnetic field)

A. $\frac{BLV}{R}$

B. $\frac{B^{2} L^{2} V^{2}}{R^{2}}$

C. $\frac{B L V}{R^{2}}$

D. $\frac{B^{2} L^{2} V^{2}}{R}$

Correct Option is (D)

From motional emf,

$e_{max} = BLV$

$\begin{array}{ll} ∴ & Heat produced = \frac{V^{2}}{r} = \frac{B^{2} L^{2} V^{2}}{R} \\ i = \frac{BLV}{R} \\ | F | = BiL and P = F \cdot V \\ ∴ & P = B (\frac{BLV}{R}) LV \\ = \frac{B^{2} L^{2} V^{2}}{R} \end{array}$

5. ⇒ (MHT CET 2023 9th May Morning Shift )

A conducting wire of length $2500 m$ is kept in east-west direction, at a height of $10 m$ from the ground. If it falls freely on the ground then the current induced in the wire is (Resistance of wire $= 25 \sqrt{2} Ω$ , acceleration due to gravity $g = 10 m / s^{2}, B_{H} = 2 \times 10^{- 5} T$ )

A. $0.2 A$

B. $0.02 A$

C. $0.01 A$

D. $2 A$

Correct Option is (B)

$\begin{aligned} We know, e = B / v \\ v^{2} = 2 g h \\ ∴ v = \sqrt{2 gh} \\ ∴ e = (2 \times 10^{- 5}) \times 2500 \times \sqrt{2 \times 10 \times 10} \\ = (2 \times 10^{- 5}) \times 25000 \times \sqrt{2} \\ = 2 \sqrt{2} \times 25 \times 10^{3} \times 10^{- 5} \\ = 50 \sqrt{2} \times 10^{- 2} V \\ ∴ I = \frac{e}{R} = \frac{50 \sqrt{2} \times 10^{- 2}}{25 \sqrt{2}} = 2 \times 10^{- 2} = 0.02 A \end{aligned}$

⇒ Electromagnetic Induction