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Ashok0210
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(a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Fig. 6.21. (b) Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, v = 10 m/s. Calculate the induced emf in the loop at the instant when x = 0.2 m. Take a = 0.1 m and assume that the loop has a large resistance.

Class 12 Physics CBSE and UP Board Electromagnetic Induction Chapter-6 Exercise 6.16 Additional Exercise NCERT Solutions for Class 12 Physics Chapter 6 Question 16

2020-2021cbse and up boardclass 12 chapter-6 physicsclass 12 physicsncert class 12 electromagnetic induction
Ashok0210
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Class 12 Physics CBSE and UP Board Electromagnetic Induction Chapter-6 Exercise 6.15 Additional Exercise NCERT Solutions for Class 12 Physics Chapter 6 Question 15

2020-2021cbse and up boardclass 12 chapter-6 physicsclass 12 physicsncert class 12 electromagnetic induction
Ashok0210
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Figure 6.20 shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T, resistance of the closed loop containing the rod = 9.0 mil. Assume the field to be uniform. (a) Suppose K is open and the rod is moved with a speed of 12 cm s_1 in the direction shown. Give the polarity and magnitude of the induced emf.(b) Is there an excess charge built up at the ends of the rods when K is open? What if K is closed? (c) With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they do experience magnetic force due to the motion of the rod. Explain. (d) What is the retarding force on the rod when K is closed? (e) How much power is required (by an external agent) to keep the rod moving at the same speed (=12 cm s_1) when K is closed? How much power is required when K is open? (f) How much power is dissipated as heat in the closed circuit? What is the source of this power? (g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?

Figure 6.20 shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T, resistance of the closed loop containing the rod = 9.0 mil. Assume the field to be uniform. (a) Suppose K is open and the rod is moved with a speed of 12 cm s_1 in the direction shown. Give the polarity and magnitude of the induced emf.(b) Is there an excess charge built up at the ends of the rods when K is open? What if K is closed? (c) With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they do experience magnetic force due to the motion of the rod. Explain. (d) What is the retarding force on the rod when K is closed? (e) How much power is required (by an external agent) to keep the rod moving at the same speed (=12 cm s_1) when K is closed? How much power is required when K is open? (f) How much power is dissipated as heat in the closed circuit? What is the source of this power? (g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?

Class 12 Physics CBSE and UP Board Electromagnetic Induction Chapter-6 Exercise 6.14 Additional Exercise NCERT Solutions for Class 12 Physics Chapter 6 Question 14

2020-2021cbse and up boardclass 12 chapter-6 physicsclass 12 physicsncert class 12 electromagnetic induction
Ashok0210
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Class 12 Physics CBSE and UP Board Electromagnetic Induction Chapter-6 Exercise 6.13 Additional Exercise NCERT Solutions for Class 12 Physics Chapter 6 Question 13

2020-2021cbse and up boardclass 12 chapter-6 physicsclass 12 physicsncert class 12 electromagnetic induction
Ashok0210
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A line charge A per unit length is lodged uniformly onto the rim of a wheel of mass M and radius R. The wheel has light non-conducting spokes and is free to rotate without friction about its axis (Fig. 6.22]. A uniform magnetic field extends over a circular region within the rim. It is given by, B = – Bo k (r ≤ a; a < R) = 0 (otherwise].What is the angular velocity of the wheel after the field is suddenly switched off?

Class 12 Physics CBSE and UP Board Electromagnetic Induction Chapter-6 Exercise 6.17 Additional Exercise NCERT Solutions for Class 12 Physics Chapter 6 Question 17

2020-2021cbse and up boardclass 12 chapter-6 physicsclass 12 physicsncert class 12 electromagnetic induction
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