Ncert Solution Class 12th Physics Chapter – 6 Electromagnetic Induction
| Textbook | NCERT |
| class | Class – 12th |
| Subject | Physics |
| Chapter | Chapter – 6 |
| Chapter Name | Electromagnetic Induction |
| Category | Class 12th Physics Question & Answer |
| Medium | English |
| Source | last doubt |
Ncert Solution Class 12th Physics Chapter – 6 Electromagnetic Induction
?Chapter – 6?
✍Electromagnetic Induction✍
?Question & Answer?
Q 1. Predict the direction of induced current in the situations described by the following figures
Answer: Lenz’s law shows the direction of the induced current in a closed loop. In the given two figures they show the direction of induced current when the North pole of a bar magnet is moved towards and away from a closed-loop respectively.
We can predict the direction induced current in different situations by using Lenz’s rule.
(i) The direction of the induced current is along qrpq.
(ii) The direction of the induced current is along prq along yzx.
(iii) The direction of the induced current is along yzxy.
(iv) The direction of the induced current is along zyxz.
(v) The direction of the induced current is along xryx.
(vi) No current is induced since the field lines are lying in the plane of the closed-loop.
Q 2. A 1.0 m long metallic rod is rotated with an angular frequency of400 rad s–1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallicring. A constant and uniform magnetic field of 0.5 T parallel to theaxis exists everywhere. Calculate the emf developed between the centre and the ring.
Answer:
Given: The length of the rod is 1 m, the angular frequency is 400 rad/sec and the magnetic field strength is 0.5 T.
The average liner velocity of the rod is given as,
v= 0+ωl 2 = ωl 2
Where, the angular frequency is ω and the length is l.
The induced emf is given as,
e=Blv =Bl( ωl 2 ) = B l 2 ω 2
Where, the magnetic field is B and the average linear velocity is v.
By substituting the given values in the above expression, we get
e= 0.5× 1 2 ×400 2 =100 V
Thus, the induced emf between the centre and the ring is 100 V.
Q 3. A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?
Answer:
Number of turns on the solenoid = 15 turns/cm = 1500 turns/m
Number of turns per unit length, n = 1500 turns
The solenoid has a small loop of area, A = 2.0 cm2 = 2 × 10−4 m2
Current carried by the solenoid changes from 2 A to 4 A.
∴
Change in current in the solenoid, di = 4 − 2 = 2 A
Change in time, dt = 0.1 s
Induced emf in the solenoid is given by Faraday’s law as:
e = dø/dt …… (1)
Where,
Ø
= Induced flux through the small loop
= BA … (ii)
B = Magnetic field
μ0ni …… (iii)
μ0 = Permeability of free space
= 4π×10−7 H/m
Hence, equation (i) reduces to:
Hence, the induced voltage in the loop is
7.54×10-6 V.
Q 4. A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of the uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s–1 in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?
Answer:
(a) Given: The length of the rectangular wire loop is 8 cm, width of the rectangular wire loop is 2 cm, the magnetic field is 0.3 T and the velocity of the loop is 1 cm/s.
The area of the loop is given as,
A=l×b
Where, the length is l and the width is b.
By substituting the given values in the above formula, we get
A=8× 10 −2 ×2× 10 −2 =16× 10 −4 m 2
The induced emf is given as,
e = Blv
Where, the magnetic field is B and the velocity is v.
By substituting the given values in the above formula, we get
e = 0.3×8× 10 −2 ×1× 10 −2 =2.4× 10 −4 V
The time taken to travel along the width is given as,
t = b v
By substituting the given values in the above formula, we get
t = 0.02 0.01 =2 s
Thus, the induced voltage is 2.4× 10 −4 V and time taken to travel along the width is 2 s.
(b) For shorter side of the loop, the induced emf is given as,
e = Bbv
By substituting the given values in the above formula, we get
e =0.3×0.02×0.01 = 0.6× 10 − 4 V
The time taken to travel along the length is given as,
t = l v
By substituting the given values in the above formula, we get
t = 0.08 0.01 =8 s
Thus, the induced voltage is 0.6× 10 −4 V and the time taken to travel along the length is 8 s.
Q 6. A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s–1 in a uniform horizontal magnetic field of magnitude 3.0 × 10–2 T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10 Ω, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from?
Answer:
Given: The radius of the circular coil is 8 cm, the numbers of turns are 20, the angular speed is 50 rad/s, the uniform magnetic field is 3× 10 −2 T and the resistance of the loop is 10 Ω.
The area of the coil is given as,
A=π r 2
Where, the radius of the circular coil is r.
By substituting the given values in the above formula, we get
A=π× ( 0.08 ) 2
The maximum induced emf is given as,
e=BANω
Where, the magnetic field strength is B, the area of the coil is A, the number of turns are N and the angular frequency is ω.
By substituting the given values in the above formula, we get
e=3× 10 −2 × ( 0.08 ) 2 ×20×50 =0.603 V
Thus, the maximum induced emf is 0.603 V.
The average emf induced in the coil over a full cycle will be zero.
The maximum current in the coil is given as,
I= e R
Where, the resistance is R.
By substituting the given values in the above formula, we get
I= 0.603 10 =0.0603 A
Thus, the maximum current in the coil is 0.0603 A.
The average power loss due to joule heating is given as,
P= eI 2
Where, the average power loss is P.
By substituting the given values in the above formula, we get
P= 0.603×0.0603 2 =0.018 W
Thus, the average power loss is 0.018 W and the induced current opposes the rotation of the coil so, an external rotor must supply the extra torque to counter this torque and keep the rotation constant. Thus, the external rotor is the source of this dissipated power.
Q 7. A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s–1, at right angles to the horizontal component of the earth’s magnetic field, 0.30 × 10–4 Wb m–2.
(a) What is the instantaneous value of the emf induced in the wire?
(b) What is the direction of the emf?
(c) Which end of the wire is at the higher electrical potential?
Answer:
Given: The length of the wire is 10 m, the falling speed of the wire is 5.0 m/s and the magnetic field strength of the earth 0.3× 10 −4 Wbm −2 .
a) The induced emf is given as,
e=Blv
Where, the length of the wire is l, the magnetic field strength is B and the falling speed of the wire is v.
By substitute the given values in the above formula, we get
e=0.3× 10 −4 ×10×5 =1.5× 10 −3 V
Thus, the instantaneous value of the emf induced in the wire is 1.5× 10 −3 V.
b) The direction of emf can be determined by the Fleming’s right hand rule. The Fleming’s right hand rule gives the direction of the induced current in a conductor moving in a magnetic field.
Thus, the wire is falling, so the direction of induced emf will be from West to East.
c) From Fleming’s right hand rule, the direction of induced emf is from West to East, so the current will flow from Eastern end to western end.
Thus, the eastern end of the wire will be at higher potential.
Q 8. Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit.
Answer:
Given: The initial current in the wire is 5.0 A, the final current in the wire is 0 A, the time taken for change is 0.1 s and the average emf is 200 V.
The average emf is given as,
e = L dI dt
Where, the self induction of the coil is L, the change in current is dI, the change in time is dt and the average emf is e.
By substituting the given values in above formula, we get
200=L ( 5−0 ) 0.1 L=4 H
Thus, the self induction of the coil is 4 H.
Q 9. A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?
Answer:
Given: The mutual inductance of a pair of coils is 1.5 H, the initial current is 0 A and final current is 20 A and the time taken for the change in current is 0.5 s.
The induced emf is given as,
e = dϕ dt (1)
Where, the change in the flux linkage with the coil is dϕ.
The equation of emf in terms of mutual inductance is given as,
e = μ dI dt (2)
Where, the mutual inductance of a pair of coils is μ, the change in current is dI and the change in time is dt.
By substituting the value of e from equation (1) to equation (2), we get
dϕ dt =μ dI dt dϕ=μdI
By substituting the given values in the above expression, we get
dϕ =1.5×20 =30 Wb
Thus, the change in flux linkage is 30 Wb.
Q 10. A jet plane is travelling towards the west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earth’s magnetic field at the location
has a magnitude of 5 × 10–4 T and the dip angle is 30°.
Answer:
Speed of the jet plane, v = 1800 km/h = 500 m/s
Wing spanof jet plane, l = 25 m
Earth’s magnetic field strength, B = 5.0 × 10−4 T
Angle of dip,
? = 300
Vertical component of Earth’s magnetic field,
BV = B sin
?
= 5 × 10−4 sin 30°
= 2.5 × 10−4 T
Voltage difference between the ends of the wing can be calculated as:
e = (BV) × l × v
= 2.5 × 10−4 × 25 × 500
= 3.125 V
Hence, the voltage difference developed between the ends of the wings is
3.125 V.
Q 11. Let us assume that the loop in question number 4 is stationary or constant but the current source which is feeding the electromagnet which is producing the magnetic field is slowly decreased. It was having an initial value of 0.3 T and the rate of reducing the field is 0.02 T / sec. If the cut is joined to form the loop having a resistance of 1.6 \Omega. Calculate how much power is lost in the form of heat? What is the source of this power?
Answer:
Q 12. A square loop of side 12 cm with its sides parallel to X and Y axes ismoved with a velocity of 8 cm s–1 in the positive x-direction in anenvironment containing a magnetic field in the positive z-direction.The field is neither uniform in space nor constant in time. It has agradient of 10–3 T cm–1along the negative x-direction (that is it increasesby 10 – 3 T cm–1as one moves in the negative x-direction), and it isdecreasing in time at the rate of 10–3 T s–1. Determine the direction andmagnitude of the induced current in the loop if its resistance is 4.50 mW.
Answer:
Given: The side of the square loop is 12 cm, the velocity of the loop is v, the gradient of the magnetic field along negative x-direction is 10 −3 Tcm -1 , the rate of decrease of the magnetic field is 10 −3 Ts -1 and the resistance of the loop is 4.50 mΩ.
The area of the loop is given as,
A= a 2
Where, the side of the square loop is a.
Since, 1 cm=0.01 m and 1 mΩ= 10 −3 Ω.
By substituting the given values in the above formula, we get
A=12× 10 −2 ×12× 10 −2 =144× 10 −4 m 2
The change in flux is given as,
dϕ dt =A× dB dx ×v
Where, the area of the square loop is A, the gradient of the magnetic field along negative x-direction and the velocity of the loop is v.
By substituting the given values in the above formula, we get
dϕ dt =144× 10 −4 × 10 −1 ×0.08 =11.52× 10 −5 Tm 2 s −1
The rate of change of the flux due to time variation is given as,
d ϕ ′ dt =A× dB dt
Where, the rate of decrease of magnetic field is dB dt .
By substituting the given values in the above expression, we get
d ϕ ′ dt =144× 10 −4 × 10 −3 =1.44× 10 −5 Tm 2 s −1
The induced emf in the loop is given as,
e = dϕ dt + d ϕ ′ dt
By substituting the given values in the above expression, we get
e =1.44× 10 −5 +11.52× 10 −5 =12.96× 10 −5 V
The induced current in the loop is given as,
i = e R
Where, the resistance of the loop is R.
By substituting the given values in the above formula, we get
i = 12.96× 10 −5 4.5× 10 −3 =2.88× 10 −2 A
Since, the flux is decreasing, then according to Lenz’s law the induced current should oppose the cause of the change. The induced current is induced in positive z direction.
Thus, the magnitude of the induced current is 2.88× 10 −2 Aand the direction of the current is positive z direction.
Q 13. It is desired to measure the magnitude of field between the poles of a powerful loud speaker magnet. A small flat search coil of area 2 cm2 with 25 closely wound turns, is positioned normal to the field direction, and then quickly snatched out of the field region. Equivalently, one can give it a quick 90° turn to bring its plane parallel to the field direction). The total charge flown in the coil (measured by a ballistic galvanometer connected to coil) is 7.5 mC. The combined resistance of the coil and the galvanometer is 0.50 Ω. Estimate the field strength of magnet.
Answer:
Area of the small flat search coil, A = 2 cm2 = 2 × 10−4 m2
Number of turns on the coil, N = 25
Total charge flowing in the coil, Q = 7.5 mC = 7.5 × 10−3 C
Total resistance of the coil and galvanometer, R = 0.50 Ω
Induced current in the coil,
I = Induced emf (e) / R …(1)
Induced emf is given as:
e = – N dϕ/dt …(2)
Where,
dϕ
= Charge in flux
Combining equations (1) and (2), we get
Initial flux through the coil, = BA
ϕ
Where,
B = Magnetic field strength
Final flux through the coil,
ϕ = 0
Integrating equation (3) on both sides, we have
But total charge,
Hence, the field strength of the magnet is 0.75 T.
Q 14. In the given figure we have a metal rod PQ which is put on the smooth rails AB and these are kept in between the two poles of permanent magnets. All these three (rod, rails and the magnetic field ) are in mutually perpendicular direction. There is a galvanometer ‘G’ connected through the rails by using a switch ‘K’.Given, Rod’s length = 15 cm , Magnetic field strength, B = 0.50 T, Resistance produced by the closed-loop = 9.0\; m\Omega. Let’s consider the field is uniform.
(i) Determine the polarity and the magnitude of the induced emf if we will keep the K open and the rod will be moved with the speed of 12 cm/s in the direction shown in the figure.
(ii) When the K was open is there any excess charge built up? Assume that K is closed then what will happen after it?
(iii) When the rod was moving uniformly and the K was open, then on the electron in the rod PQ there was no net force even though they did not experience any magnetic field because of the motion of the rod. Explain.
(iv) After closing the K, calculate the retarding force.
(v) When the K will be closed calculate the total external power which will be required to keep moving the rod with the same speed ( 12 cm/s)? and also calculate the power required when K will be closed.
(vi) What would be the power loss ( in form of heat) when the circuit is closed? What would be the source of this power?
(vii) Calculate the emf induced in the moving rod if the direction of the magnetic field is changed from perpendicular to parallel to the rails?
Answer:
(a) The length of the given coil is 15 cm.
l=15 cm =0.15 m
Magnetic field strength, B=0.50 T.
Resistance of the loop is R=9.0 mΩ or 9× 10 −3 Ω.
The velocity at which the rod is moved is given as v=12 cm/s or 0.12 m/s
The formula to calculate the induced emf is,
e=Bvl
Substitute values in the above formula.
e=0.5×0.12×0.15 =0.009 V =9× 10 −3 V =9 mV
The Lenz’s law determines the polarity of the induced emf. The induced emf must oppose the cause of the change. With reference to the Fleming’s right hand rule, the induced current should flow from P to Q, so the end P shows positive polarity and Q shows negative polarity.
(b) When key K is open, charge is developed at the ends of the rods and on closing the key, this charge flows in the form of current.
Therefore, yes, the excess charge is maintained due to continuous flow of the current in the circuit on closing key K.
(c) When the rod starts to move in the magnetic field, excess change is developed at the ends of the rod which develops an electric force. This force cancels the magnetic force on the electron, thus the net force on the electrons is zero.
(d) The formula to calculate the force on the conductor moving in magnetic field is,
F=IBl
Where,
I is the induced current.
B is the magnetic field strength.
l is the length of the conductor.
Induced current is calculated as,
I= e R = 9× 10 −3 9× 10 −3 =1 A
The length of the rod is 0.15 m.
Magnetic field strength, B=0.50 T.
Substitute the value in the above formula,
F=IBl =1×0.5×0.15 =75× 10 −3 N
Thus, the retarding force on the rod is 75× 10 −3 N.
(e) The power from external source is required to overcome the resisting force of 75× 10 −3 N.
The velocity at which the rod moves is given as v=12 cm/s or 0.12 m/s
The power is calculated as,
P=Fv =75× 10 −3 ×0.12 =9× 10 −3 W =9 mW
When key K is open, there is no current flow in the rod, thus no power is required.
(f) The formula to calculate the power, in terms of current i is,
Power dissipated as heat= I 2 R = ( 1 ) 2 ×9× 10 −3 =9× 10 −3 =9 mW
Thus, the power dissipated as heat is 9 mW and the source of this power is external source.
(g) If the magnetic field is parallel to the rod, then there will not be any induced emf as the coil would not cut any flux, moving in the parallel direction.
Thus, the induced emf in the coil will be zero.
Q 15. An air-cored solenoid with length 30 cm, area of cross-section 25 cm2and number of turns 500, carries a current of 2.5 A. The current issuddenly switched off in a brief time of 10–3 s. How much is the averageback emf induced across the ends of the open switch in the circuit?Ignore the variation in magnetic field near the ends of the solenoid.
Answer:
The given length of the solenoid, l=30 cm or 0.3 m.
Cross section area of the solenoid, A=25 cm 2 or 25× 10 −4 m 2 .
Number of turns, N=500.
Current in the solenoid, I=2.5 A.
Time for which the current flows, t= 10 −3 s.
The formula for average back emf is,
e= dϕ dt (1)
Change in flux is given by, dϕ=NAB and magnetic field strength is B= μ 0 NI l .
Substitute the values in equation (1).
e= μ 0 N 2 IA lt = 4π× 10 −7 × ( 500 ) 2 ×2.5×25× 10 −4 0.3× 10 −3 = 1.96× 10 −3 0.3× 10 −3 =6.5 V
Thus, the average back emf in the solenoid is 6.5 V.
Q 16. (i) We are given a long straight wire and a square loop of given size (refer to figure). Find out an expression for the mutual inductance between both.
(ii) Now, consider that we passed an electric current through the straight wire of 50 A, and the loop is then moved to the right with constant velocity, v = 10 m/s.Find the emf induced in the loop at an instant where x = 0.2 m. Take a = 0.01 m and assume that the loop has a large resistance.
Answer: (i) Take a small element dy in the loop at a distance y from the long straight wire (as shown in the given figure).
Magnetic flux
dy, dϕ=BdA
associated with element
Where,
dA = Area of element dy = a dy
B = Magnetic field at distance y
= μ0I / 2лу
I = Current in the wire
μ0
= Permeability of free space = 4π × 10−7 T m A−1
y tends from x to
a×x
(ii) Emf induced in the loop, e = B’av
= (μ0I / 2л×) av
Given,
I = 50 A
x = 0.2 m
a = 0.1 m
v = 10 m/s
Q 17. A line charge λ per unit length is lodged uniformly onto the rim of a wheel of mass ? and radius R. The wheel has light non-conducting spokes and is free to rotate without friction about its axis (Fig.). A uniform magnetic field extends over a circular region within the rim. It is given by, B=–B0^k(r≤a;a<R)=0 (otherwise)What is the angular velocity of the wheel after the field is suddenly switched off ?
Answer: