Question

A metallic rod of length L is rotated at an angular speed ω normal to a uniform magnetic field B. The emf induced in the rod is​

Answer 1

Answer:

Solution : 

If θ is the angle traced by the free and in time t, then area swept out, ltnbrgt A=πl2×(θ2π)=12l2θ 

Magnetic flux linked , ϕ=B(12l2θ)coc0∘[∴ϕBAcosθ] 

ϕ=12Bl2θ 

According to Faraday's laws of electromagnetic induction, 

Induced emf, e=dθdt=12Bl2dθdt=12Bl2ω 

∴ Induced current, I=eR=12Bl2ωR=Bl2ω2R.

Explanation:

thank you

Answer 2

Given:

A metallic rod of length L is rotated at an angular speed ω normal to a uniform magnetic field B.

To find:

EMF induced in rod.

Calculation:

Due to rotation of rod, the flux passing:

$\therefore\: \phi=B\times A$

$=>\: \phi=B\times (\dfrac{\pi{L}^{2}}{2\pi}\times \theta)$

$=>\: \phi=\dfrac{B{L}^{2}\theta}{2}$

According to Faraday's Law of Electromagnetic Induction, Let induced EMF be E

$\therefore\: E = \dfrac{d\phi}{dt}$

$=>\: E = \dfrac{d(\dfrac{B{L}^{2}\theta}{2})}{dt}$

$=>\: E = \dfrac{B{L}^{2}}{2}\times \dfrac{d\theta}{dt}$

$=>\: E = \dfrac{B{L}^{2}}{2}\times \omega$

$=>\: E = \dfrac{B{L}^{2}\omega}{2}$

So, final answer is:

$\boxed{\bf{E_{net} =\dfrac{B{L}^{2}\omega}{2} }}$

• This type of induced EMF due to movement of rod is called Motional EMF.