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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Electromagnetic Induction

NTA Abhyas 2020 · Physics · Electromagnetic Induction

A metal bar AB can slide on two parallel thick metallic rails...

A metal bar ABAB can slide on two parallel thick metallic rails separated by a distance ll. A resistance RR and an inductance LL are connected to the rails as shown in the figure. A long straight wire, carrying a constant current I0{I}_{0} is placed in the plane of the rails as shown. The bar ABAB is held at rest at a distance x0{x}_{0} from the long wire. At t=0t = 0, it is made to slide on the rails away from the wire.

A metal bar AB can slide on two parallel thick metallic rails separated by a distance l . A resistance R and an

It is observed that at timet=Tt = T, the metal bar ABAB is at a distance of 2x02{x}_{0} from the long wire and the resistance RR carries a current i1{i}_{1}. The net charge that has flown through resistance RR from t=0t = 0 to t=Tt = T is

Options

  1. A\dfrac{1}{R} \left[\dfrac{{μ}_{0}l}{2\pi}\text\ln\left(2\right) - L{i}_{1}\right]
  2. B\dfrac{1}{R} \left[\dfrac{{μ}_{0}{I}_{0}l}{π}\text\ln\left(2\right) - L{i}_{1}\right]
  3. C\dfrac{1}{R} \left[\dfrac{{μ}_{0}{I}_{0}l}{2\pi}\text\ln\left(2\right) - L{i}_{1}\right]
  4. D\dfrac{1}{R} \left[\dfrac{{μ}_{0}{I}_{0}}{2\pi}\text\ln\left(2\right) - L{i}_{1}\right]

Correct answer

C. \dfrac{1}{R} \left[\dfrac{{μ}_{0}{I}_{0}l}{2\pi}\text\ln\left(2\right) - L{i}_{1}\right]

Step-by-step solution

Let v be velocity of AB when AB is at a distance x from the wire,

eAB=B.ℓ.v=μ0I02πxI.v=dϕdt{e}_{AB} = \text{B} \text{.} ℓ \text{.} \text{v} = \dfrac{{μ}_{0} {I}_{0}}{2 \pi x} \text{I} \text{.} \text{v} = \dfrac{\text{d} ϕ}{\text{d} \text{t}}

For circuit Loop,

dϕdt−iR−Ldidt=0\dfrac{\text{d} ϕ}{\text{d} \text{t}} - \text{i} \text{R} - \text{L} \dfrac{\text{d} \text{i}}{\text{d} \text{t}} = 0

dϕdt−dqdtR−Ldidt=0\dfrac{\text{d} ϕ}{\text{d} \text{t}} - \dfrac{\text{d} \text{q}}{\text{d} \text{t}} \text{R} - \text{L} \dfrac{\text{d} \text{i}}{\text{d} \text{t}} = 0

⇒dq=dϕ−LdiR⇒ \text{d} \text{q} = \dfrac{\text{d} ϕ - \text{L} \text{d} \text{i}}{\text{R}}

\Deltaq=\Deltaϕ−L\DeltaIR\text{\Delta} \text{q} = \dfrac{\text{\Delta} ϕ - \text{L} \text{\Delta} \text{I}}{\text{R}}

=1R[(ϕ1−ϕ2)−Li1]= \dfrac{1}{\text{R}} [ \left({ϕ}_{1} - {ϕ}_{2}\right) - \text{L} {i}_{1} ]

\Deltaq=1R[μ0ℓI02\piln⁡(2)−Li1]\text{\Delta}\text{q} = \dfrac{1}{R} [ \dfrac{{μ}_{0}ℓ{I}_{0}}{2\text{\pi}}\ln \left(2\right) - \text{L}{i}_{1} ]
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