Answer to Question #175930 in Electric Circuits for Joshua Musyoki

Question

A conductor of length 𝐿 moves at a steady speed vat right angles to a uniform magnetic field of flux density 𝐵. Show that the e.m.f. 𝐸 across the ends of the conductor is given by the 

equation 𝐸 = 𝐵𝑙𝑣. 

Answer 1

Let the velocity of the rod be $v \space m/s$

Magnetic flux density = B T/m²

Length of rod = $l$ m

Any change in magnetic flux induces an emf opposing that change—a process known as induction. Motion is one of the major causes of induction.

Charges moving in a magnetic field experience the magnetic force $F=qvBsin\theta$ , which moves opposite charges in opposite directions.

From Faraday's law,

Magnitude of emf induced along the moving rod = $N\dfrac{\Delta\Phi }{\Delta t}$ (1)

As the number of turns in this coil is equal to one

$\therefore$ N=1

Flux, $\Phi=BAcos\theta$

Since, magnetic field is perpendicular to the area of cross-section

$\theta=0\degree,\space cos0\degree=1$

$\Delta\Phi=\Delta (BA)\\\Rightarrow\Delta\Phi=B\Delta A\\\Rightarrow\Delta\Phi=Bl\Delta x$

Substituting the value of N and $\Delta\Phi$ in equation (1),

we get,

emf $=Bl\dfrac{\Delta x}{\Delta t}$

Since $\dfrac{\Delta x}{\Delta t}=v,$ the velocity of the rod

$\therefore \space emf=Blv$ ($B,l$ and $v$ are perpendicular to each other)

The above expression gives the motional emf.