Download the app

Questions  

Figure shows an Amperian path ABCDA. Part ABC is in vertical plane PSTU while part CDA is in horizontal plane POQRS. Direction of circulation along the path is shown by an arrow near point B and at D.

B·d for this path according to Ampere’s law will be

Remember concepts with our Masterclasses.

80k Users
60 mins Expert Faculty Ask Questions
a
I1-I2+I3μ0
b
-I1+I2μ0
c
I3μ0
d
I1+I2μ0

Ready to Test Your Skills?

Check Your Performance Today with our Free Mock Tests used by Toppers!

detailed solution

Correct option is D

∮B→·d→ℓ=μ0I1+I3+I2-I3=μ0I1+I2  [Since for the given direction of circulation I3 enteringatPSTU  is positive while I3 at PQRS is negative] Alternative solution: ∮ABCDAB→·dl→=∮ABCAB→·dl→+∮CDACB→·dl→ =μ0I1+I3+μ0I2-I3=μ0I1+I2

Talk to our academic expert!

+91

Are you a Sri Chaitanya student?

ctaimg

Create Your Own Test
Your Topic, Your Difficulty, Your Pace


Similar Questions

A cylindrical wire of radius R is carrying uniformly distributed current i over its cross-section. If a circular loop of radius r is taken as amperian loop, then the variation value of  B·d over this loop with radius ‘r” of loop will be best represented by


phone icon
whats app icon