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

NTA Abhyas 2020 · Physics · Electromagnetic Induction

An equilateral triangular loop ADC having some resistance is pulled...

An equilateral triangular loop ADCADC having some resistance is pulled with a constant velocity vv out of a uniform magnetic field directed into the screen. At time t=0t = 0, side DCDC of the loop is at edge of the magnetic field. The induced current \left(i\right)~versus& time \left(t\right) graph will be as

An equilateral triangular loop ADC having some resistance is pulled with a constant velocity v out of a uniform

Options

  1. AIf Δ = | cases x - 2 & 2x - 3 & 3x - 4 2x - 3 & 3x - 4 & 4x - 5 3x - 5 & 5x - 8 & 10x - 17 cases | = A x ^ 3 + B x ^ 2
  2. BIf Δ = | cases x - 2 & 2x - 3 & 3x - 4 2x - 3 & 3x - 4 & 4x - 5 3x - 5 & 5x - 8 & 10x - 17 cases | = A x ^ 3 + B x ^ 2
  3. CIf Δ = | cases x - 2 & 2x - 3 & 3x - 4 2x - 3 & 3x - 4 & 4x - 5 3x - 5 & 5x - 8 & 10x - 17 cases | = A x ^ 3 + B x ^ 2
  4. DIf Δ = | cases x - 2 & 2x - 3 & 3x - 4 2x - 3 & 3x - 4 & 4x - 5 3x - 5 & 5x - 8 & 10x - 17 cases | = A x ^ 3 + B x ^ 2

Correct answer

B. If Δ = | cases x - 2 & 2x - 3 & 3x - 4 2x - 3 & 3x - 4 & 4x - 5 3x - 5 & 5x - 8 & 10x - 17 cases | = A x ^ 3 + B x ^ 2

Step-by-step solution

Let 2a be the side of the triangle and b the length AE.

An equilateral triangular loop ADC having some resistance is pulled with a constant velocity v out of a uniform

AHAE=GHEC\dfrac{\text{AH}}{\text{AE}} = \dfrac{\text{GH}}{\text{EC}}

∴GH=(AHAE)EC∴ \text{GH} = \left(\dfrac{\text{AH}}{\text{AE}}\right) \text{EC}

or GH=(b−vt)b⋅a\text{GH} = \dfrac{\left(\text{b} - \text{vt}\right)}{\text{b}} \cdot \text{a}

=a−(abvt)<br/><br/>= \text{a} - \left(\dfrac{\text{a}}{\text{b}} \text{vt}\right)<br /><br />∴ FG=2GH=2[a−abvt]\text{FG} =2 \text{GH} =2 \left[\text{a} - \dfrac{\text{a}}{\text{b}} \text{vt}\right]

Induced emf e=Bv(FG)=2Bv(a−abvt)<br/><br/>\text{e} = \text{Bv} \left(\text{FG}\right) =2 \text{Bv}\left(\text{a} - \dfrac{\text{a}}{\text{b}} \text{vt}\right)<br /><br />∴Inducedcurrenti=eR=2BvR[a−abvt]Induced current \text{i} = \dfrac{\text{e}}{\text{R}} = \dfrac{2 \text{Bv}}{\text{R}}\left[\text{a} - \dfrac{\text{a}}{\text{b}} \text{vt}\right]

or& & i = k1 - k2t

Thus, i-t graph is a straight line with negative slope and positive intercept.
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