Question

An armature coil consists of 20 turns of wire, each of area A = 0.09m2 and total resistance 15.0 . It rotates in a
megnetic field of 0.5T at a constant frequency of . Calculate the value of (i) maximum (ii) average
induced emf produced in the coil

Answer 1

Answer:

( i ) The maximum value of induce emf is 43 volt

( i i ) The average value of induce emf is 0 volt .

Explanation:

Given as :

Number of coil turn = N = 20

Area of wire = A = 0.09 m²

resistance = R = 15 ohm

The value of magnetic field = B = 0.5 Tesla

ω  = $\dfrac{150}{3.14}$  = 47.7

According to question

(i) from  law of electromagnetism

emf = - N $\dfrac{\partial \Phi }{\partial t}$

And  ∅  = BA cosωt

i.e emf = - N $\dfrac{\partial BA cos\omega t}{\partial t}$

or, emf = - NBAω Sinωt

For maximum emf  , ωt = 90°

∴  emf = - NBAω Sin 90°

or, $emf_m_a_x$ = -  NBAω            ( Sin 90° = 1 )

Or, $emf_m_a_x$ = - 20 × 0.09 × 0.5 × 47.7

i.e $emf_m_a_x$ = 42.99 ≈ 43 volt

Hence, The maximum value of induce emf is 43 volt . Answer

( i i ) For average induced emf

emf = - N $\dfrac{\partial BA cos\omega t}{\partial t}$

or, emf = - NBAω Sinωt

For average emf  , ωt = 0° for period of oscillation

∴  emf = - NBAω Sin 0°

Or, $emf_a_v_g$ = 0

Hence, The average value of induce emf is 0 volt . Answer