Question 325: Electromagnetic Induction

Question

A metal wire of length $$2500 \mathrm{~m}$$ is kept in east-west direction, at a height of $$10 \mathrm{~m}$$ from the ground. If it falls freely on the ground then the current induced in the wire is (Resistance of wire $$=25 \sqrt{2} \Omega, \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$$ and Earth's horizontal component of magnetic field $$\left.\mathrm{B}_{\mathrm{H}}=2 	imes 10^{-5} \mathrm{~T}ight)$$

Accepted Answer

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 :
$EMF = B 	imes v 	imes l$
Where :
$B$ is the magnetic field (which is given as $B_H = 2 	imes 10^{-5} T$)
$v$ 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 :
$v^2 = u^2 + 2gh$
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 :
$EMF = 2 	imes 10^{-5} T 	imes 14.14 m/s 	imes 2500 m$
$EMF = 0.707 V$
Determine the current using Ohm's Law:
$I = \frac{EMF}{R}$
Where :
$I$ is the current
$EMF$ is the voltage induced (0.707 V)
$R$ is the resistance (which is $25\sqrt{2} \Omega$)
Given that $\sqrt{2}$ is approximately 1.414 :
$R = 25(1.414)$
$R = 35.35 \Omega$
Now, plugging in the values to find the current :
$I = \frac{0.707 V}{35.35 \Omega}$
$I \approx 0.02 A$
Therefore, the answer is :
Option B
0.02 A

Answer

0.2 A

Answer

0.02 A

Answer

0.01 A

Answer

2 A

Electromagnetic Induction

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