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

Nov 15, 2017 16:51 IST
NCERT Exemplar Solutions 12th Physics - Chapter 6: Electromagnetic Induction

NCERT Exemplar Solutions for Class 12 Physics - Chapter 6: Electromagnetic Induction are available here. In this article, you will get solutions for long answer type questions i.e., question number 22 to 32. These questions are very important for Class 12 Physics board exam and other competitive exams. Solutions for Multiple Choice Questions with Single Correct Answer (MCQ 1) and Multiple Correct Answers (MCQ 2), Very Short Answer Type Questions (VSA) & Short Answer Type Questions (SA) are already available in previous articles.

Solutions for long answer type questions of NCERT Exemplar 12th Physics - Chapter 6: Electromagnetic Induction are given below

Question 22:

A magnetic field B= B sin (wt) kˆ covers a large region where a wire AB slides smoothly over two parallel conductors separated by a distance d (figure). The wires are in the x-y plane. The wire AB (of length d) has resistance R and the parallel wires have negligible resistance. If AB is moving with velocity v, what is the current in the circuit. What is the force needed to keep the wire moving at constant velocity?

Solution 22:

Question 23:

A conducting wire XY of mass m and negligible resistance slides smoothly on two parallel conducting wires as shown in figure. The closed circuit has a resistance R due to AC. AB and CD are perfect conductors. There is a magnetic field B = B(t) kˆ

(i) Write down equation for the acceleration of the wire XY

(ii) If B is independent of time, obtain v (t), assuming v (0) = u

(iii) For (ii), show that the decrease in kinetic energy of XY equals the heat lost in.

Solution 23:

Question 24:

ODBAC is a fixed rectangular conductor of negligible resistance (CO is not connected) and OP is a conductor which rotates clockwise with an angular velocity w (figure). The entire system is in a uniform magnetic field B whose direction is along the normal to the surface of the rectangular conductor ABDC. The conductor OP is in electric contact with ABDC. The rotating conductor has a resistance of l per unit length. Find the current in the rotating conductor, as it rotates by 180°.

Solution 24:

Question 25:

Consider an infinitely long wire carrying a current I (t), with dI/dt = λ= constant. Find the current produced in the rectangular Loop of wire ABCD if its resistance is R (as show in the figure given below).

Solution 25:

Question 26:

A rectangular Loop of wire ABCD is kept close to an infinitely long wire carrying a current I (t) = I (1 - t/T) for 0 ≤ t T and I (0) = 0 for t >T (figure.). Find the total charge passing through a given point in the loop, in time T. The resistance of the loop is R.

Solution 26:

Question 27:

A magnetic field B is confined to a region ra and points out of the paper (the z-axis), r = 0 being the centre of the circular region. A charged ring (charge = Q) of radius b, b > a and mass m lies in the x-y plane with its centre at the origin. The ring is free to rotate and is at rest. The magnetic field is brought to zero in time At. Find the angular velocity co of the ring after the field vanishes.

Solution 27:

Question 28:

A rod of mass m and resistance R slides smoothly over two parallel perfectly conducting wires kept sloping at an angle 0 with respect to the horizontal (figure). The circuit is closed through a perfect conductor at the top. There is a constant magnetic field B along the vertical direction. If the rod is initially at rest, find the velocity of the rod as a function of time.

Solution 28:

Question 29:

Find the current in the sliding rod AB (resistance = R) for the arrangement shown in figure. B is constant and is out of the paper. Parallel wires have no resistance, v is constant. Switch S is closed at time t = 0.

Solution 29:

Question 30:

Find the current in the sliding rod AB (resistance = R) for the arrangement shown in figure. B is constant and is out of the paper. Parallel wires have no resistance, v is constant. Switch S is closed at time t = 0.

Solution 30:

Question 31:

A metallic ring of mass m and radius l (ring being horizontal) is falling under gravity in a region having a magnetic field. If z is the vertical direction, the z-component of magnetic field is Bz = B (1 + l z). If R is the resistance of the ring and if the ring falls with a velocity v, find the energy lost in the resistance. If the ring has reached a constant velocity, use the conservation of energy to determine v in terms of m, B, l and acceleration due to gravity g.

Solution 31:

Question 32:

A long solenoid S has n turns per meter, with diameter a. At the centre of this coil, we place a smaller coil of N turns and diameter b (where b < a). If the current in the solenoid increases linearly, with time, what is the induced emf appearing in the smaller coil. Plot graph showing nature of variation in emf, if current varies as a function of mt2 + C.

Solution 32:

NCERT Solutions for CBSE Class 12 Physics: All Chapters

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