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★★★★★
68% · 34/50 Questions
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Start your 14-day free trial to unlock the full solution →The back (self-induced) emf is , so it tracks the slope of the i–t graph. The steepest segment is AB, so the back emf is greatest for . Comparing slopes gives , and .
Concept
For a solenoid (inductor) the self-induced back emf is
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 to : .
- AB: from to : .
- BC: from to : .
- After C (): current constant at , so .
Where the back emf is maximum
The largest magnitude of slope is on AB, so the back emf is maximum throughout the interval . …
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