Question

A metallic rod of length 'l' is moved perpendicular to its length with velocity 'v' in a magnetic field 'B' acting perpendicular to the plane in which rod moves . Derive the expression for the induced emf​

Answer 1

Then a force will act on conductor By faraday's laws of electromagnetic induction,

V = \dfrac{d\phi}{dt}V=dtdϕ

V = B\dfrac{dA}{dt}V=BdtdA

V = BlvV=Blv as dA/dt = lvdA/dt=lv

This is the induced emf across the ends of the rod.

(b)

We have the force on the free electrons from Q to P as \vec F_b=\vec {qv}\times \vec BFb=qv×B

where vv is the velocity of the rod as well as of the free electrons inside it, BB is the uniform magnetic field. The free electrons will move towards P and positive charge will appear at Q. An electrostatic field is developed from Q to P in the wire which exerts a force \vec F_e=q\vec EFe=qE on each electron. The charge keeps on gathering until

\vec F_b=\vec F_eFb=Fe and the resultant force on each electron is zero.

|q\vec v\times \vec B|=|q\vec E|∣qv×B∣=∣qE∣

\Rightarrow vB=E⇒vB=E

The potential difference between Q and P is then 

V=El=vBlV=El=vBl

which is maintained by the magnetic force of the moving electron producing an emf, e=Bvle=Bvl

Answer 2

Then a force will act on conductor By faraday's laws of electromagnetic induction,

V = \dfrac{d\phi}{dt}V=dtdϕ

V = B\dfrac{dA}{dt}V=BdtdA

V = BlvV=Blv as dA/dt = lvdA/dt=lv

This is the induced emf across the ends of the rod.

(b)

We have the force on the free electrons from Q to P as \vec F_b=\vec {qv}\times \vec BFb=qv×B

where vv is the velocity of the rod as well as of the free electrons inside it, BB is the uniform magnetic field. The free electrons will move towards P and positive charge will appear at Q. An electrostatic field is developed from Q to P in the wire which exerts a force \vec F_e=q\vec EFe=qE on each electron. The charge keeps on gathering until

\vec F_b=\vec F_eFb=Fe and the resultant force on each electron is zero.

|q\vec v\times \vec B|=|q\vec E|∣qv×B∣=∣qE∣

\Rightarrow vB=E⇒vB=E

The potential difference between Q and P is then 

V=El=vBlV=El=vBl

which is maintained by the magnetic force of the moving electron producing an emf, e=Bvle=Bvl

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