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NCERT Exemplar · Q16

Q.The current i through a solenoid varies with time t along three straight-line segments OA, AB and BC. Starting from the origin O at (t = 0, i = 0), the current rises linearly to point A at (t = 5 s, i = +1 A); it then falls linearly to point B at (t = 10 s, i = −2 A); and finally rises linearly to point C at (t = 30 s, i = 0), after which it stays constant at zero. For which interval of time is the magnitude of the back (self-induced) electromotive force a maximum? If the back emf at t = 3 s equals e, find the back emf at t = 7 s, t = 15 s and t = 40 s.

Rajasthan RbseSubjective· 3mImportance★★★★★
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The back (self-induced) emf is ε=−Ldidt\varepsilon=-L\dfrac{di}{dt}, so it tracks the slope of the i–t graph. The steepest segment is AB, so the back emf is greatest for 5 s<t<10 s5\text{ s}<t<10\text{ s}. Comparing slopes gives ε(7)=−3e\varepsilon(7)=-3e, ε(15)=e/2\varepsilon(15)=e/2 and ε(40)=0\varepsilon(40)=0.

Concept

For a solenoid (inductor) the self-induced back emf is

ε=−Ldidt,\varepsilon=-L\frac{di}{dt},

proportional in magnitude to the slope of the current–time graph. On a straight segment the slope, and hence the back emf, is constant.

Slopes of the three segments

  • OA: from (0,0)(0,0) to (5 s,+1 A)(5\text{ s},+1\text{ A}): didt=15=+0.2 A s−1\dfrac{di}{dt}=\dfrac{1}{5}=+0.2\text{ A s}^{-1}.
  • AB: from (5 s,+1 A)(5\text{ s},+1\text{ A}) to (10 s,−2 A)(10\text{ s},-2\text{ A}): didt=−2−110−5=−0.6 A s−1\dfrac{di}{dt}=\dfrac{-2-1}{10-5}=-0.6\text{ A s}^{-1}.
  • BC: from (10 s,−2 A)(10\text{ s},-2\text{ A}) to (30 s,0)(30\text{ s},0): didt=0−(−2)30−10=+0.1 A s−1\dfrac{di}{dt}=\dfrac{0-(-2)}{30-10}=+0.1\text{ A s}^{-1}.
  • After C (t>30 st>30\text{ s}): current constant at 00, so didt=0\dfrac{di}{dt}=0.

Where the back emf is maximum

The largest magnitude of slope is 0.6 A s−10.6\text{ A s}^{-1} on AB, so the back emf is maximum throughout the interval 5 s<t<10 s5\text{ s}<t<10\text{ s}. …

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