Question 53: Electromagnetic Induction

Question

A 20 m long uniform copper wire held horizontally is allowed to fall under the gravity $\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)$ through a uniform horizontal magnetic field of 0.5 Gauss perpendicular to the length of the wire. The induced EMF across the wire when it travells a vertical distance of 200 m is $\_\_\_\_$ mV .

Accepted Answer

The wire falls from rest (initial velocity $u=0$ ) under gravity.
Using the third equation of motion its velocity after falling a distance $h=200 \mathrm{~m}$ is,
$$ v^2=u^2+2 g h $$
$$\Rightarrow $$ $v^2=0^2+2(10)(200)$
$$\Rightarrow $$ $\mathrm{v}^2=4000$
$$\Rightarrow $$ $\mathrm{v}=\sqrt{4000} \mathrm{~m} / \mathrm{s}$
$$\Rightarrow $$ $$ \mathrm{v}=20 \sqrt{10} \mathrm{~m} / \mathrm{s} $$
The formula for motional EMF induced in a conductor moving perpendicular to a magnetic field is :
$$ \mathrm{E}=\mathrm{Bvl} $$
Where $B$ is magnetic field strength, $v$ is the velocity of the conductor and $l$ is the length of the conductor
$$ E=\left(5 	imes 10^{-5}ight) 	imes(20 \sqrt{10}) 	imes(20) $$
$$\Rightarrow $$ $E=2000 	imes 10^{-5} 	imes \sqrt{10}$
$$\Rightarrow $$ $E=2 	imes 10^{-2} 	imes \sqrt{10}$ Volts
$$\Rightarrow $$ $E=\left(2 	imes 10^{-2} \sqrt{10}ight) 	imes 10^3 \mathrm{mV}$
$$\Rightarrow $$ $\mathrm{E}=20 \sqrt{10} \mathrm{mV}$

Answer

$200 \sqrt{10}$

Answer

$0.2 \sqrt{10}$

Answer

$20 \sqrt{10}$

Answer

$2 \sqrt{10}$

Electromagnetic Induction

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