6. ⇒ (MHT CET 2021 21th September Morning Shift )

A wire of length ' $L$ '; having resistance ' $R$ ' falls from a height ' $ℓ$ ' in earth's horizontal magnetic field ' $B$ '. The current through the wire is ( $g =$ acceleration due to gravity)

A. $\frac{BL \sqrt{2 g ℓ}}{R}$

B. $\frac{BL \sqrt{2 g ℓ}}{R^{2}}$

C. $\frac{2 BLg ℓ}{R^{2}}$

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

Correct Option is (A)

If the wire falls through a height $h$ , the velocity acquired by it is $v = \sqrt{2 g ℓ}$

The emf induced in the wire

$\begin{aligned} e = BLv = BL \sqrt{2 g ℓ} \\ ∴ Current I & = \frac{e}{R} = \frac{BL \sqrt{2 g ℓ}}{R} \end{aligned}$

7. ⇒ (MHT CET 2021 20th September Evening Shift )

A metal 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 $= 25 \sqrt{2} Ω, g = 10 m / s^{2}$ and Earth's horizontal component of magnetic field $B_{H} = 2 \times 10^{- 5} T)$

A. 0.2 A

B. 0.02 A

C. 0.01 A

D. 2 A

Correct Option is (B)

To find the current induced in the wire, we first need to determine the change in magnetic flux as the wire falls.

Determine the EMF (voltage) induced in the wire :

The voltage induced (EMF) when a conductor moves in a magnetic field can be given by Faraday's law of electromagnetic induction :
$E M F = B \times v \times l$
Where :

$B$ is the magnetic field (which is given as $B_{H} = 2 \times 10^{- 5} T$ )
$v$ is the velocity of the wire
$l$ is the length of the wire (which is 2500 m)

When the wire falls freely under gravity, it will accelerate due to gravitational force. The velocity $v$ of the wire just before hitting the ground can be found using the kinematic equation :

$v^{2} = u^{2} + 2 g h$
Where :

$u$ is the initial velocity (which is 0, since the wire starts from rest)
$g$ is the acceleration due to gravity (10 m/s^2)
$h$ is the height (10 m)

Plugging in the given values :

$v^{2} = 0 + 2 (10 m / s^{2}) (10 m)$
$v^{2} = 200$
$v = \sqrt{200}$
$v = 14.14 m / s$

Now, using Faraday's law :

$E M F = 2 \times 10^{- 5} T \times 14.14 m / s \times 2500 m$
$E M F = 0.707 V$

Determine the current using Ohm's Law:

$I = \frac{E M F}{R}$

Where :
$I$ is the current
$E M F$ is the voltage induced (0.707 V)
$R$ is the resistance (which is $25 \sqrt{2} Ω$ )

Given that $\sqrt{2}$ is approximately 1.414 :
$R = 25 (1.414)$
$R = 35.35 Ω$

Now, plugging in the values to find the current :
$I = \frac{0.707 V}{35.35 Ω}$
$I \approx 0.02 A$

Therefore, the answer is :
Option B
0.02 A

8. ⇒ (MHT CET 2021 20th September Morning Shift )

A straight conductor of length 0.6 M is moved with a speed of 10 ms $^{- 1}$ perpendicular to magnetic field of induction 1.2 weber m $^{- 2}$ . The induced e.m.f. across the conductor is

A. 6 V

B. 7.2 V

C. 0.72 V

D. 12 V

Correct Option is (B)

The induced emf is given by

e = B $l$ v = $1.2 \times 0.6 \times 10 = 7.2$ V

⇒ Electromagnetic Induction