CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

A small circular loop of radius l=1 cm is kept inside a larger circular loop of radius r such that both are coplanar and concentric. If the current in larger loop is as shown in the graph, then current induced in the smaller loop any time is represented as In at any instant. Take anticlockwise current as positive. List -1 contains time instants, and List-2 contains the values for In. Match them. Take μ0π2Rr=1.

List -1List -2(I)0 sec(p)25μA(II)13sec(Q)12.5μA(III)32sec(R)0μA(IV)53sec(S)6.25μA(T)12.5μA(U)25μA

A
IP,IIR,IIIU,IVP
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
IU,IIR,IIIP,IVU
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
IP,IIR,IIIP,IVP
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
IU,IIR,IIIU,IVU
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D IU,IIR,IIIU,IVU
ϕ=BA=μ0i2rπl2
=dϕdt
=μ0π2rl2(didt)
In=R
In=μ0π2rRl2(didt)=104(didt)=100(didt)μA
at t=0,didt=11614=14In=25μA clockwise
So at t=0, In=25μA
at t=13sec, didt=0In=0
at t=32sec, didt=14In=25μA clockwise
At t=32, In=25μA
at t=53sec, didt=14In=25μA clockwise.
At t=53, In=25μA
So, IU,IIR,IIIU,IVU

flag
Suggest Corrections
thumbs-up
0
mid-banner-image
mid-banner-image
similar_icon
Similar questions
View More
similar_icon
Related Videos
thumbnail
lock
PHYSICS
Watch in App