Question

A coil of wire of area 0.2m^(2) containing 1000 turns is placed in a magnetic field of induction √2T. At time t=0 the axis of the coil and the direction of magnetic field are along z-direction.The coil is rotated by an angle 45° in 2s about y -axis. Assuming the constant angular speed the average e.m.f induced in the coil is approximately
Ans :41.4V

Answer 1

Given quantities:

number of turns(n)=1000

Area of the coil(A)= 0.2 square meter

Magnetic field(B)= $\sqrt{2}$ Tesla

Angle($\delta\theta$) rotated by 45$\degree$ in time($\delta{t}$) 2 sec with constant angular velocity ($\omega$) .

Detailed solution

formula for flux is...

$$\phi=\vec{B}.\vec{A}= B A \cos{\theta}=B A \cos{\omega t}\\ emf=-\bigg(\cfrac{d\phi}{dt}\bigg)\\ emf=\omega n B A \sin{\omega t}\\ <emf>=\cfrac{\int_0^2{(emf) dt}}{\int_0^2{dt}}\\ <emf>=-\cfrac{nBA}{2}\bigg|\cos(\omega t)\bigg|_0^2\\ <emf>=-\cfrac{nBA}{2}[\cos{2\omega }- \cos{0}].....eq[1]\\$$

and

$$\delta\theta=\omega\delta t\\ \cfrac{\pi}{4}=\omega \times 2\\ \omega =\cfrac{\pi}{8}\\$$

Now putting the value of $\omega$ in eq[1] and we get...

$$<emf>=-\cfrac{nBA}{2}\bigg[\cos(\cfrac{\pi}{4})-1\bigg]\\$$

putting the values of n, B and A, we get..

$$<emf>=\cfrac{1000\times \sqrt{2}\times0.2 }{2}\bigg[\cfrac{\sqrt{2}-1}{\sqrt{2}}\bigg]\\ =100(\sqrt2-1)\\ <emf>=41.4 volt.....Ans$$

average e.m.f induced in the coil is 41.4 volt.