CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

The flux of magnetic field through a closed conducting loop of resistance 0.4 changes with time according to the equation ϕ = 0.20t2 + 0.40t + 0.60 where t is time in seconds. Find (i) the induced emf at t = 2s. (ii) the average induced emf in t = 0 to t = 5 s. (iii) change passed through the loop in t = 0 to t = 5s. (iv) average current in time interval t = 0 to t = 5s (v) heat produced in t = 0 to t = 5s.

Open in App
Solution

i) E=dϕdt=0.4t+0.4=0.8+0.4=1.2Volt

ii) Eav=ΔϕΔt=ϕ5ϕ050=[(0.2×25+0.4×5+0.6)0.6]5=1.4Volts

iii) As E=IR

or I=ER

dqdt=ER

dq=ERdt

Q0dq=1R50(0.4t+0.4)dt

Q=10.4[0.2t2+0.4t]50=(0.2×25)+(0.4×5)0.4=17.5 Columbs

iv) Iavg=ΔQΔt=17.55=3.5A

v) As dH=I2Rdt=V2Rdt

As V=0.4t+0.4

H0dH=50(0.4t+0.4)20.4dt

H=0.4[t33+t2+t]50=863 Joule

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Dimensional Analysis
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon