NCERT Solution Physics Class 12 Chapter 6 Electromagnetic Induction

NCERT Solution Physics Class 12 Chapter 6 Electromagnetic Induction

NCERT Solution Physics Class 12 Chapter 6 Electromagnetic Induction all questions and answers. Physics Class 12 6th Chapter Electromagnetic Induction exercise solution and experts answer. As one of online learning platforms, we (netex.) are excited to offer the NCERT Solution Physics Class 12 Chapter 6. This solution is designed to help students who are looking to brush up on their physics concepts on Chapter 6 Electromagnetic Induction.

 

6.1 Predict the direction of induced current in the situations described by the following Figs. 6.18(a) to (f ).

ANSWER-

We know that when magnetic flux linked with coil changes then current is induced in the coil And direction of this current is such that to oppose the source which changes flux linked with coil.

(a)From figure (a) south pole of the magnet is approaching towards the coil hence current will induce such that magnetic field will have south pole on left side so that it oppose the magnet approaching the coil as south pole repel the south pole hence according to right hand thumb rule induced current should be along qrpq.

(b)From figure (b)  south pole of the magnet is approaching towards the coil hence current will induce such that induced magnetic field will have south pole on left side so that it oppose the magnet approaching the coil as south pole repel the south pole hence according to right hand thumb rule induced current should be along prqp.

(c) from figure c, current is flowing through circuit in clockwise direction when observed from the coil xyz and produces magnetic field such that having south pole towards coil xyz hence to oppose the magnetism produced by circuit current will induced in coil xyz with south pole on circult side to oppose the south pole produced by circuit . hence by right hand thumb rule induced current will be along xyzx.

(d)from figure d, current is flowing through circuit in anticlockwise direction when observed from the coil xyz and produces magnetic field such that having north pole towards coil xyz hence to oppose the magnetism produced by circuit, current will induced in coil xyz with north pole towards circuit side to oppose the north pole produced by circuit . Hence by right hand thumb rule induced current will be along zyxz.

(e)The direction of the induced current is along xryx.

(f) from figure f  , current carrying wire is vertical hence produces the magnetic field line in horizontal plane and loop is also is placed in the same plane hence no flux linked with the coil changes hence no current is induced in the loop

 

6.2.) Use Lenz’s law to determine the direction of induced current in the situations described by Fig. 6.19: (a) A wire of irregular shape turning into a circular shape; (b) A circular loop being deformed into a narrow straight wire.

ANSWER-

(a) From figure (a) we observe that area of wire in irregular shape is comparatively less than the circular area.so when a wire of irregular shape turning into circular shape flux linked with the coil increases as area in contact increases thereby increasing the number of magnetic lines passing through the circular coil increases. As magnetic field is coming out of the plane of paper. Hence current is induced to oppose increase in flux in outward direction such that it flows along abcd.

(b) from figure b we observe that while converting the circular loop into straight wire the area linked with the magnetic field is decreases hence flux coming out of the wire will also decrease. This decrease in the flux results into current in the wire such that to oppose the decreasing flux hence according to Fleming’s right hand rule current direction will be a’d’c’b’.

 

6.3) 

 

6.6)

 

6.11)

 

6.12) 

 

In case you are missed :- NCERT Solution for Alternating Current

 

6.13)

 

6.14.) 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 mΩ. 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) 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?

ANSWER-

(a)Suppose K is open and the rod is moved with a speed of 12 cm s–1 in the direction shown in figure then according to left hand rule we can force is acting on rod from Q  to P which indicates that P will have positive polarity while Q will have negative polarity. To find the magnitude we know that induced emf and given by,

ⅇ = BLv

given that,

B = 0.50 T , Length of the rod = 15 cm = 0.15m, v =12 cm s–1 = 0.12 m/s

ⅇ = 0.50 x 0.15 x 0.12

ⅇ = 9 x 10-3 v.

(b)Excess charges are build up at the ends of the rod. When k is closed then these excess charges get path to flow and current starts flowing.

(c)With K open there is magnetic force on the the electron in the direction from Q to P. but due to this build up a electric field is set up in the rod  having electrical force opposite to magnetic force hence they balances each other and hence , there is no net force on the electrons in the rod PQ .

(d) the magnitude of retarding force is given by,

F = IBL

∴ F = IBL

∴ F = E/R X BL

∴ F = 9×10-3/9×10-3 x 0.5 x 0.15

∴ F = 0.075N

 

(e)When key k is closed,

Power required to keep the rod moving at the speed(12 cm/s) is

p = force x velocity

∴ p = 0.075 ´ 0.12

∴ p = 0.009 W.

When k is open no power is required.

 

(f) Power dissipated as a heat is given by,

 P = I2R

(g) current will be induced in the moving rod only if the direction of motion of the rod is perpendicular to the direction of magnetic field. If the magnetic field is parallel to the rails instead of being perpendicular then no current will be induced.

 

6.15) 

 

6.16 (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.

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.

ANSWER-

(a)Take the element of length dy at a distance y from the wire as shown below.

Magnetic flux associated with strip of length dyis given by,

∴ dΦ = B dA        

Taking integration over the entire area with limits y = x to  y = x +a

This is expression for mutual inductance between a long straight wire and a square loop of side a.

(b) given that, x = 0.2 m. Take a = 0.1 m

Emf induced in the square coil at a distance x due to current through wire is

 

6.17) 

In case you are missed :- NCERT Solution for Electromagnetic Waves

For more update, follow this page NCERT Solutions

Updated: April 15, 2023 — 3:21 pm

Leave a Reply

Your email address will not be published. Required fields are marked *