wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A long solenoid of radius R carries a time (t) dependent current I(t)=I0t(1t). A ring of radius 2R is placed coaxially near its middle. During the time instant 0t1, the induced current (IR) and the inducedEMF(VR) in the ring changes as:


A

Direction of IR remains unchanged and VR is maximum att=0.5

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

Direction ofIR remains unchanged and VR is zero att=0.25

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

At t=0.5 direction of IR reverses and VR is zero

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D

At t=0.25 direction of IR reverses and VR is maximum

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C

At t=0.5 direction of IR reverses and VR is zero


Step 1: Given data

Current I(t)=I0t(1t)

The radius of the ring =2R

The radius of the solenoid =R

Step 2: Find out the flux, ϕ

Field due to solenoid near the middle, B=µoNI (Where, N=number of turns per unit length)

Flux, ϕ=B.A [Where A=πR2(cross-sectional area)]

=µoNI.πR2=µoNIot(1-t).πR2

Step 3: Find out the relation between the induced current (IR) and the inducedEMF(VR) in the ring and draw the diagram

Induced emf, VR=dϕ/dt [By Lenz’s law]

=-dµoNIot(1-t).πR2dt=πμoIoNR2(12t)

As, V=IR And we can see VR is zero for

(1-2t)=0t=(1/2)s

VR is negative for t<1/2 and increasing linearly

VR is positive for t>1/2 and increasing linearly

The same will be true for current, I[As,VR]

Hence, option (C) is correct.


flag
Suggest Corrections
thumbs-up
2
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Faraday’s Law of Induction
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon