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

If the resistances of values R1 and R2 are connected on both ends as shown in figure, current I, flowing through resistance R1 is given by
224740_a2d571ca0f864375be6f73bae8334662.png

A
BlR2(v1r2v2r1)R1R2(r1+r2)+r2r1(R1+R2)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
BlR2(v1r2+v2r1)R1R2(r1+r2)+r2r1(R1+R2)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
BlR2(v1r2v2r1)R1R2(r1r2)+r2r1(R1+R2)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
BlR2(v1r2v2r1)R1R2(r1+r2)r2r1(R1+R2)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
E
none
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A BlR2(v1r2v2r1)R1R2(r1+r2)+r2r1(R1+R2)
As the conducting rods are moving in magnetic field with velocity v in opposite direction, thus an EMF is induced in each rod according to the Lenz law.
EMF induced in rod MN E1=Bv1l .........(a)
EMF induced in rod MN E2=Bv2l .......(b)

Combining the EMFs of the two rods, 1req=1r1+1r2 req=r1r2r1+r2 .......(1)
Also equivalent EMF Eeq=E2r2reqE1r1req
Putting the values and on solving we get, Eeq=E2r1E1r2r1+r2 ............(2)
From Kirchhoff current law i=i1+i2
Using Nodal analysis EeqVreq=V0R1+V0R2

V=EeqreqR1R2req[(R1+R2)req+R1R2] .............(3)

i1=VR1
From (1), (2) and (3), we get i1= (E2r1E1r2)r1r2R2r1r2[(R1+R2)r1r2+(r1+r2)R1R2]

Using (a) and (b), |i1|= |v2r1v1r2|BlR2[(R1+R2)r1r2+(r1+r2)R1R2]

480361_224740_ans_e454e7caebdf462ea533425f9fc4930e.png

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