A coil of 200 turns and area 0.20 m^2 is rot | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The magnetic flux linked with the coil is given by $\phi = NAB \cos(\omega t)$.The induced electromotive force $(EMF)$ is $\varepsilon = -\frac{d\

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) The magnetic flux linked with the coil is given by $\phi = NAB \cos(\omega t)$.The induced electromotive force $(EMF)$ is $\varepsilon = -\frac{d\phi}{dt} = NAB\omega \sin(\omega t)$.The maximum voltage generated is $\varepsilon_{\max} = NAB\omega$.Given:Number of turns $N = 200$Area $A = 0.20 \ m^2$Magnetic field $B = 0.01 \ T$Frequency $f = 0.5 \ rev/s$,so angular frequency $\omega = 2\pi f = 2\pi(0.5) = \pi \ rad/s$.Substituting the values:$\varepsilon_{\max} = 200 	imes 0.20 	imes 0.01 	imes \pi$$\varepsilon_{\max} = 40 	imes 0.01 	imes \pi = 0.4\pi = \frac{4\pi}{10} = \frac{2\pi}{5} \ V$.Comparing this with $\frac{2\pi}{\beta}$,we get $\beta = 5$.

Similar Questions

When a magnet is pushed in and out of a circular coil $C$ connected to a very sensitive galvanometer $G$ as shown in the adjoining diagram with a frequency $\nu$,then

$A$ circular coil of resistance $R$,area $A$,and number of turns $N$ is rotated about its vertical diameter with angular speed $\omega$ in a uniform magnetic field of magnitude $B$. The average power dissipated in a complete cycle is:

The phase difference between the magnetic flux linked with a coil rotating in a uniform magnetic field and the induced $e.m.f.$ produced in it is

Discuss the principle and construction of an $A.C.$ generator (dynamo). Derive the formula for the induced $emf$ in an $A.C.$ generator.

In an $AC$ generator,a coil with $N$ turns,all of the same area $A$ and total resistance $R$,rotates with frequency $\omega$ in a magnetic field $B$. The maximum value of emf generated in the coil is

Vedclass Test Series

Exam Paper Generator

Online Exam Module