NCERT Exemplar Solutions for Class 12 Physics - Chapter 6: Electromagnetic Induction (SA)

NCERT Exemplar Solutions for Class 12 Physics (Chapter 7 – Alternating Current) are available here. In this article, you will get solutions from question number 6.17 to question number 6.21. These are short answer type question.

Solutions of multiple choice questions with single correct answer (MCQ I) and multiple correct answers (MCQ II) are already available.

These questions are important for CBSE Class 12 Physics board exam and various competitive exams.

NCERT Exemplar Solutions for CBSE Class 12 Physics, Chapter 6 (from question number 6.17 to 6.21) are given below:

Question 6.17:

Solution 6.17:

Question 6.18:

Consider a closed Loop C in a magnetic field (figure). The flux passing through the loop is defined by choosing a surface whose edge coincides with the loop and using the formula f = B1dA1, B2dA2…Now, if we choose two different surfaces S1 and S2 having C as their edge, would we get the same answer for flux. Justify your answer

Solution 6.18:

One gets the same answer for flux. Flux can be thought of as the number of magnetic field lines passing through the surface (we draw dN = B Δ A lines in an area Δ A ⊥to B), As lines of of B cannot end or start in space (they form closed loops) number of lines passing through surface S1 must be the same as the number of lines passing through the surface S2.

Question 6.19:

Find the current in the wire for the configuration shown in figure. Wire PQ has negligible resistance. B, the magnetic field is coming out of the paper. q is a fixed angle made by PQ travelling smoothly over two conducting parallel wires separated by a distance d.

Solution 6.19:

Motional electric field E along the dotted line CD (⊥ to both v and B and along v × B) = vB

E.M.F. along PQ = (length PQ) × (Field along PQ)

E.M.F. along PQ = (length PQ) × (Field along PQ) = (d/cosθ) × vB cosθ = dvB.

So, I = dvB/R and is independent of q.

Question 6.20:

A (current versus time) graph of the current passing through a solenoid is shown in figure. For which time is the back electromotive force (u) a maximum. If the back emf at t = 3 s is e, find the back emf at t = 7 s, 15s and 40 s. OA, AB and BC are straight line segments.

Solution 6.20:

We know that, back emf, e = L (dI/dt), so e will be maximum when dI/dt is maximum.

From the graph, we can see that, maximum rate of change of current is in between interval AB. Therefore, maximum back emf will be obtained between 5s < t < 10s.

For, 0 s < t < 5 s,

dI/dt = 1/5. Now, at t = 3 seconds, e = LdI/dt = L/5


For 5 s < t < 10 s

dI/dt = ‒ 3/5 and at t = 7 seconds e = ‒3e

For 10 s < t < 30 s

dI/dt = 1/10 and at t = 15 seconds e = e/2

For, t > 30 s

dI/dt = 1/10 and e = 0

Question 6.21:

There are two coils A and B separated by some distance. If a current of 2 A flows through A, a magnetic flux of 10‒2 Wb passes through B (no current through B). If no current passes through A and a current of 1 A passes through B, what is the flux through A?

Solution 6.21:

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