Question 20: Electromagnetic Induction

Question

A metallic rod of $$2 \mathrm{~m}$$ length is rotated with a frequency $$100 \mathrm{~Hz}$$ about an axis passing through the centre of the circular ring of radius $$2 \mathrm{~m}$$. A constant magnetic field $$2 \mathrm{~T}$$ is applied parallel to the axis and perpendicular to the length of the rod. The emf developed across the ends of the rod is :

Accepted Answer

To determine the electromotive force (emf) developed across the ends of the rotating rod, we can use the concept of motional emf generated in a rotating conductor in a magnetic field.
The formula for the motional emf in a rotating rod is given by:
$$ \varepsilon = \frac{1}{2} B \omega L^2 $$
where:
$$B$$ is the magnetic field strength
$$\omega$$ is the angular velocity
$$L$$ is the length of the rotating rod
First, let's find the angular velocity $$\omega$$. The frequency $$f$$ is given as $$100 \, 	ext{Hz}$$. The relationship between angular velocity $$\omega$$ and frequency $$f$$ is:
$$ \omega = 2 \pi f $$
Plugging in the values:
$$ \omega = 2 \pi \cdot 100 = 200 \pi \, 	ext{rad/s} $$
The length of the rod $$L$$ is given as $$2 \, 	ext{m}$$ and the magnetic field strength $$B$$ is given as $$2 \, 	ext{T}$$.
Now, substituting these values into the motional emf formula:
$$ \varepsilon = \frac{1}{2} \cdot 2 \cdot 200 \pi \cdot (2)^2 $$
Simplifying this expression:
$$ \varepsilon = \frac{1}{2} \cdot 2 \cdot 200 \pi \cdot 4 $$
$$ \varepsilon = 800 \pi \, 	ext{V} $$
Therefore, the emf developed across the ends of the rod is:
Option A
: 800 $$ \pi $$ volt

Answer

800 $$\pi$$ volt

Answer

1600 $$\pi$$ volt

Answer

1600 volt

Answer

400 $$\pi$$ volt

Electromagnetic Induction

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