KVPY SX SB Exam 2015 Question Paper

Question : 35 of 120

Marks: +1, -0

A metallic ring of radius a and resistance R is held fixed with its axis along a spatially uniform magnetic field whose magnitude is

B_0 \sin (ωt)

. Neglect gravity. Then,

Solution:

Emf

= {-dϕ}/{dt}⇒ε=-d/{dt}(BA)

⇒ε=-{AdB}/{dt}

⇒

ε=-A\ B_0 ω\cos ωt ⇒i=ε/R

i=-{B_0ωA}/R\cos\ ωt

current oscillates with "ω".
Heating loss =

i^2R

H ∝ i^2\;\;\;[i={B_0ω(πa^2)}/R\cos \ ωt]

H ∝ B_0 ^ 2 ω^ 2 a ^4

Force on

d l

length

|F| = \Bid l

|F| = B_0 \sin \ ωt ({B_0ω πa^2 }/R)\cos \ ωt.dl

Force per unit length =

|F|/{dl}={B_0^2ω πa^2 }/R

sin ωt cos ωt
Force per unit length

∝ B_0^2

Net force on ring will be zero.

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