Question

A helicopter has blades of length 4.0 m rotating at 3.0 rev/s in a horizontal plane.If the vertical component of the Earth’s magnetic field is 6.5 105T,what is the induced emf between the tip of a blade and the hub?

Answer 1

Answer:

Induced emf, $\epsilon=9.79\times 10^7\ volts$

Explanation:

Given that,

Length of the helicopter, l = 4 m

Angular speed of the helicopter, $\omega=3\ rev/s=18.84\ rad/s$

The vertical component of the Earth’s magnetic field is, $B=6.5\times 10^5\ T$

We need to find the induced emf between the tip of a blade and the hub. The induced emf in terms of angular velocity of an rotating object is given by :

$\epsilon=\dfrac{1}{2}B\omega l^2$

$\epsilon=\dfrac{1}{2}\times 6.5\times 10^5\times 18.84\times (4)^2$

$\epsilon=9.79\times 10^7\ volts$

So, the induced emf between the tip of a blade and the hub is $\epsilon=9.79\times 10^7\ volts$. Hence, this is the required solution.