Name: Production of Induced EMF by Rotating the Loop
Uploaded: 2019-07-19
Duration: 13 min 16 s
Description: Learn how the magnetic flux can be altered by rotating a coil in a magnetic field such that the angle between the magnetic field and the area vector changes.

Question

A fan blade of length

0.5 m

rotates uniformly at right angles to a uniform magnetic field of

5 mT

parallel to the axis of rotation. If the fan makes

1500

revolutions per minute how much potential difference is developed between the ends of the blade during rotation?

Accepted Answer

Here, radius of the circle traced by the fan blade, $R = 0.5 m$

Frequency of rotation, $n = 1500 r e v o l u t i o n s p e r m i n u t e = 25 r e v o l u t i o n s p e r s e c o n d$

Magnetic field strength, $B = 5 mT$

In the figure above, the rotating fan blade goes from $O P$ to $O Q$ in $t s$ tracing $∠ P O Q = θ$

Let $e$ be the potential difference developed between the ends of the blade during rotation.

$T h e p o t e n t i a l d i f f e r e n c e a c r o s s t h e f a n b l a d e = e m f b e t w e e n c e n t r e a n d e n d o f t h e f a n b l a d e$

$\Rightarrow e = B \times r a t e o f c h a n g e o f a r e a o f l o o p$

$\Rightarrow e = B \times \frac{d (a r e a o f l o o p P O Q)}{d t}$

$\Rightarrow e = B \times \frac{d (π R^{2} \times \frac{θ}{2 π})}{d t}$

$\Rightarrow e = \frac{1}{2} B R^{2} \frac{d θ}{d t}$

$\Rightarrow e = \frac{1}{2} B R^{2} ω = \frac{1}{2} B R^{2} (2 π n)$

$\Rightarrow e = π n B R^{2} = \frac{22}{7} \times 25 \times 5 \times 10^{- 3} \times {(0.5)}^{2} = 98.125 \times 10^{- 3} V$

$∴ e = 0.0981 V$

So the potential difference between the ends of the blade during rotation is $0.0981 V$ .

A fan blade of length $0.5 m$ rotates uniformly at right angles to a uniform magnetic field of $5 mT$ parallel to the axis of rotation. If the fan makes $1500$ revolutions per minute how much potential difference is developed between the ends of the blade during rotation?

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