__Motional e.m.f __: If a conductor is moving in a magnetic field , electrons inside it experience a force $ \displaystyle \vec{F} = -e (\vec{v}\times \vec{B}) $ and accumulate at the end of the conductor. Very soon , an electric field is established.

Eventually component of magnetic force along the conductor length is balanced by the electric field force and the drifting of electrons stops and an emf is established. Now ,

$ \displaystyle \xi = -\int \vec{E}.\vec{dl} = \int (\vec{v}\times \vec{B}).\vec{dl} $

This is the general expression for induced emf in a conducting wire . If v , B & l are mutually perpendicular to each other then ξ = Blv

Example : A copper rod of length ‘l’ rotates at an angular velocity ‘ ω ‘ in a uniform magnetic field B as shown in figure. What is the induced emf across its ends?

Solution: The rod is supposed to be the combination of a number of infinitesimal elements. Speed of each element is different. Consider an element at a distance l from O. Speed of this element is ωl, and is perpendicular to its length.

$ \displaystyle \xi = \int B v (dl )$

As v = ωl

$ \displaystyle \xi = \omega B \int_{0}^{L} ldl $

$ \displaystyle \xi = \frac{1}{2} B \omega l^2 $

‘O’ turns out to be ‘+’ and P the ‘-‘ ve terminal.

Exercise : An air plane with 20m wing spread is flying at 250 m/s straight along south parallel to the earth surface. The earth’s magnetic field has a horizontal component of 2 × 10^{-5} Wb/m^{2} and the dip angle is 60°. Calculate the induced emf. between the wing tips.

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