Ncert Solutions Physics Class 12th
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11.4 Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW.
(a) Find the energy and momentum of each photon in the light beam,
(b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and
(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?
11.4 Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW.
(a) Find the energy and momentum of each photon in the light beam,
(b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and
(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?
11.4 Wavelength of the monochromatic light, = 632.8 m
Power emitted by laser, P = 9.42 mW = 9.42 W
Planck’s constant, h = 6.626 Js
Speed of light, c = 3 m/s
Mass of hydrogen atom, m = 1.66 kg
The energy of each photon is given as, &nb
Upvote 
11.3 The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?
11.3 The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?
11.3 Photoelectric cut-off voltage, = 1.5 V
Maximum kinetic energy of the photoelectrons emitted is given as
, where e = charge of an electron = 1.6 C
K = 1.6 = 2.4 J
Hence, maximum kinetic energy of the photoelectrons emitted is 2.4 .
Upvote 11.2 The work function of cesium metal is 2.14 eV. When light of frequency 6 Hz is incident on the metal surface, photoemission of electrons occurs. What is the
(a) Maximum kinetic energy of the emitted electrons,
(b) Stopping potential, and
(c) Maximum speed of the emitted photoelectrons?
11.2 The work function of cesium metal is 2.14 eV. When light of frequency 6 Hz is incident on the metal surface, photoemission of electrons occurs. What is the
(a) Maximum kinetic energy of the emitted electrons,
(b) Stopping potential, and
(c) Maximum speed of the emitted photoelectrons?
11.2 Work function of cesium metal, = 2.14 eV
Frequency of light, = 6 Hz
The maximum kinetic energy is given by the photoelectric effect, = ,
Where = Planck’s constant = 6.626 Js, 1 eV = 1.602 J
= =
Upvote 6.17 A line charge l 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 = – B0 k (r a; a < R)
= 0 (otherwise)
What is the angular velocity of the wheel after the field is suddenly switched off?
6.17 A line charge l 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 = – B0 k (r a; a < R)
= 0 (otherwise)
What is the angular velocity of the wheel after the field is suddenly switched off?
6.17 Line charge per unit length = = =
Where r = distance of the point within the wheel
Mass of the wheel = M
Radius of the wheel = R
Magnetic field, =
At a distance r, the magnetic force is balanced by the centripetal force. i.e.
BQ = , w
Upvote 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.
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.
6.16 Let us take a small element dy in the loop, at a distance y from the long straight wire.
Magnetic flux associated with element dy, d = BdA, where
dA = Area of the element dy = a dy
Magnetic flux at distance y, B = , where
I = current in the wire
= Permeability of fr
Upvote 6.15 An air-cored solenoid with length 30 cm, area of cross-section 25 cm2 and number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10–3 s. How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid.
6.15 An air-cored solenoid with length 30 cm, area of cross-section 25 cm2 and number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10–3 s. How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid.
6.15 Length of the solenoid, l = 30 cm = 0.3 m
Area of cross-section, A = 25 = 25
Number of turns on the solenoid, N = 500
Current in the solenoid, I = 2.5 A
Current flowing time, t = s
We know average back emf, e = …………………. (1)
Where Change in flux
Upvote 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 (=12cm 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?
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 (=12cm 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?
6.14 Length of the rod, l = 15 cm = 0.15 m
Magnetic field strength, B = 0.5 T
Resistance of the closed loop, R = 9 mΩ = 9 Ω
Induced emf = 9 mV
Polarity of the induced emf is such that end P shows positive while end Q shows negative ends.
Speed of the rod, v = 12 cm/s = 0.12 m/s
Induced emf is g
Upvote 6.13 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.
6.13 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.
6.13 Area of the coil, A = 2 = 2
Number of turns, n = 25
Total charge flowing in the coil, Q = 7.5 mC = 7.5 C
Total resistance of the coil and galvanometer, R = 0.50 Ω
We know, Induced current in the coil, I = ……… (1)
Induced emf is given by the equation, e =&
Upvote 6.12 A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm s–1 in the positive x-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of 10–3 T cm–1 along the negative x-direction (that is it increases by 10 – 3 T cm–1 as one moves in the negative x-direction), and it is decreasing in time at the rate of 10–3 T s–1. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 mΩ.
6.12 A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm s–1 in the positive x-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of 10–3 T cm–1 along the negative x-direction (that is it increases by 10 – 3 T cm–1 as one moves in the negative x-direction), and it is decreasing in time at the rate of 10–3 T s–1. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 mΩ.
6.12 Side of the square loop, s = 12 cm = 0.12 m
Area of the square loop, A = 0.12
Velocity of the loop, v = 8 cm /s = 0.08 m/s
Gradient of the magnetic field along negative x – direction.
= T/cm = T/m
Rate of decrease of magnetic field
= T/s
Resistance of
Upvote 6.11 Suppose the loop in Exercise 6.4 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.3 T at the rate of 0.02 T s–1. If the cut is joined and the loop has a resistance of 1.6 Ω, how much power is dissipated by the loop as heat? What is the source of this power?
6.11 Suppose the loop in Exercise 6.4 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.3 T at the rate of 0.02 T s–1. If the cut is joined and the loop has a resistance of 1.6 Ω, how much power is dissipated by the loop as heat? What is the source of this power?
6.11 The area of the rectangular coil, A = 8 16 = 16
Initial value of the magnetic field, = 0.3 T
Rate of decrease of the magnetic field, = 0.02 T/s
From the relation of induced emf e = , where is the change in the flux linka
Upvote