Resistance of the resistor, R = 100 Ω, Supply voltage, V = 220 V, Frequency ,ν = 50 Hz Ans (a). The rms value of current in the circuit is given as, I =V/R =220/100 = 2.20 A Ans (b). The net power consumed over a full cycle is given as : P = VI = 220 x 2.2 = 484 W
Resistance of the resistor, R = 100 Ω,
Supply voltage, V = 220 V,
Frequency ,ν = 50 Hz
Ans (a).
The rms value of current in the circuit is given as,
I =V/R =220/100 = 2.20 A
Ans (b).
The net power consumed over a full cycle is given as :
Line charge per unit length = λ = Total Charge/Length = Q/2πr Where, r = Distance of the point within the wheel Mass of the wheel = M Radius of the wheel = R Magnetic field, B = -B0 k At distance r, the magnetic force is balanced by the centripetal force i.e., BQv = Mv²/r Where, v = linear velocityRead more
Line charge per unit length = λ = Total Charge/Length = Q/2πr
Where, r = Distance of the point within the wheel
Mass of the wheel = M
Radius of the wheel = R
Magnetic field, B = -B0 k
At distance r, the magnetic force is balanced by the centripetal force i.e.,
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 associated with element dy, dφ = BdA Where, dA = Area of element dy = a dy B = Magnetic field at distance y =μ0I/2πy I = Current in the wire μ0= Permeability of free space =Read more
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 associated with element dy, dφ = BdA
Where, dA = Area of element dy = a dy
B = Magnetic field at distance y
=μ0I/2πy
I = Current in the wire
μ0= Permeability of free space = 4π x 10⁻⁷
Therefore ,dφ = (μ0Ia/2π) x (dy/y)
φ = μ0Ia/2π ⌠ dy/y { NOTE- “⌠” is sign of integration}
v tends from x to a + .x.
Therefore ,φ = μ0Ia/2π x⌠a+x dy/y
=μ0Ia/2π [loge y]xa+x
=μ0Ia/2π [loge(a+x)/x]
For mutual inductance M. the flux is given as: φ = MI
Length of the solenoid, 1 = 30 cm = 0.3 m Area of cross-section, A = 25 cm2 = 25 x 10-4 m2 Number of turns on the solenoid, N = 500 Current in the solenoid, I = 2.5 A Current flows for time, t = 10⁻3 s Average back emf, e = dφ/dt---------------------Eq-1 Where, dφ = Change in flux = NAB ------------Read more
Length of the solenoid, 1 = 30 cm = 0.3 m
Area of cross-section, A = 25 cm2 = 25 x 10-4 m2
Number of turns on the solenoid, N = 500
Current in the solenoid, I = 2.5 A
Current flows for time, t = 10⁻3 s
Average back emf, e = dφ/dt———————Eq-1
Where, dφ = Change in flux = NAB ——————-Eq-2
B = Magnetic field strength =μ0NI/l——————–Eq-3
Where, μ0= Permeability of free space = 4π x 10-7 T m A-1
Using equations (2) and (3) in equation (1), we get
e = μ0N²IA/lt
= [(4π x 10-7) x (500)² x 2.5 x 25 x 10-4]/(0.3 x 10⁻3)
=6.5 V
Hence, the average back emf induced in the solenoid is 6.5 V.
Length of the rod, l = 15 cm = 0.15 m Magnetic field strength, B = 0.50 T Resistance of the closed loop, R = 9 mΩ = 9 x 10⁻³ Ω Ans (a). 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 InducedRead more
Length of the rod, l = 15 cm = 0.15 m
Magnetic field strength, B = 0.50 T
Resistance of the closed loop, R = 9 mΩ = 9 x 10⁻³ Ω
Ans (a).
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 given as:
e = Bvl
= 0.5 x 0.12 x 0.15 = 9 x 10-3 v = 9 mV
The polarity of the induced emf is such that end P shows positive while end Q shows negative ends.
Ans (b).
Yes; when key K is closed, excess charge is maintained by the continuous flow of current.
When key K is open, there is excess charge built up at both ends of the rods.
When key K is closed, excess charge is maintained by the continuous flow of current.
Ans (c).
Magnetic force is cancelled by the electric force set-up due to the excess charge of opposite nature at both ends of the rod.
There is no net force on the electrons in rod PQ when key K is open and the rod is moving uniformly. This is because magnetic force is cancelled by the electric force set-up due to the excess charge of opposite nature at both ends of the rods.
Ans (d).
Retarding force exerted on the rod, F = IBl Where,
I = Current flowing through the rod
= e/R = (9 x 10-3 )/(9 x 10-3 ) =1 A
Therefore, F = 1 x 0.5 x 0.15 =75 x 10-3 N
Ans (e).
9 mW; no power is expended when key K is open.
Speed of the rod, v = 12 cm/s = 0.12 m/s
Hence, power is given as:
P = Fv
= 75 x 10-3 x0.12 = 9×10-3 W = 9 mW
When key K is open, no power is expended.
Ans (f).
9 mW; power is provided by an external agent.
Power dissipated as heat = I2 R = (l)2 x 9x 10-3 = 9 mW.
The source of this power is an external agent.
Ans (g).
Zero
In this case, no emf is induced in the coil because the motion of the rod does not cut across the field lines.
A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply. (a) What is the rms value of current in the circuit? (b) What is the net power consumed over a full cycle?
Resistance of the resistor, R = 100 Ω, Supply voltage, V = 220 V, Frequency ,ν = 50 Hz Ans (a). The rms value of current in the circuit is given as, I =V/R =220/100 = 2.20 A Ans (b). The net power consumed over a full cycle is given as : P = VI = 220 x 2.2 = 484 W
Resistance of the resistor, R = 100 Ω,
Supply voltage, V = 220 V,
Frequency ,ν = 50 Hz
Ans (a).
The rms value of current in the circuit is given as,
I =V/R =220/100 = 2.20 A
Ans (b).
The net power consumed over a full cycle is given as :
P = VI = 220 x 2.2 = 484 W
See lessA 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?
Line charge per unit length = λ = Total Charge/Length = Q/2πr Where, r = Distance of the point within the wheel Mass of the wheel = M Radius of the wheel = R Magnetic field, B = -B0 k At distance r, the magnetic force is balanced by the centripetal force i.e., BQv = Mv²/r Where, v = linear velocityRead more
Line charge per unit length = λ = Total Charge/Length = Q/2πr
Where, r = Distance of the point within the wheel
Mass of the wheel = M
Radius of the wheel = R
Magnetic field, B = -B0 k
At distance r, the magnetic force is balanced by the centripetal force i.e.,
BQv = Mv²/r
Where,
v = linear velocity of the wheel
Therefore, B2πrλ = Mv/r
v =B2πλr²/M
Therefore ,Angular velocity , ω =v/R = B2πλr²/MR
For r ≤ a ≤ R, we get
ω = -(2πB0a²λ/MR) k^
See less(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.
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 associated with element dy, dφ = BdA Where, dA = Area of element dy = a dy B = Magnetic field at distance y =μ0I/2πy I = Current in the wire μ0= Permeability of free space =Read more
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 associated with element dy, dφ = BdA
Where, dA = Area of element dy = a dy
B = Magnetic field at distance y
=μ0I/2πy
I = Current in the wire
μ0= Permeability of free space = 4π x 10⁻⁷
Therefore ,dφ = (μ0Ia/2π) x (dy/y)
φ = μ0Ia/2π ⌠ dy/y { NOTE- “⌠” is sign of integration}
v tends from x to a + .x.
Therefore ,φ = μ0Ia/2π x⌠a+x dy/y
=μ0Ia/2π [loge y]xa+x
=μ0Ia/2π [loge(a+x)/x]
For mutual inductance M. the flux is given as: φ = MI
Therefore ,MI = μ0Ia/2π [loge(a/x+1)]
M=μ0a/2π [loge(a/x+1)]
Ans (b).
Emf induced in the loop, e = B’av
= (μ0I/2π)x av
Given, I = 50 A
x = 0.2 m
a = 0.1 m
v= 10 m/s
e = (4π x 10⁻⁷ x 50 x 0.1 x 10 )/(2π x 0.2)
e = 5 x 10⁻⁵ V
See lessAn 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 10⁻³s. 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.
Length of the solenoid, 1 = 30 cm = 0.3 m Area of cross-section, A = 25 cm2 = 25 x 10-4 m2 Number of turns on the solenoid, N = 500 Current in the solenoid, I = 2.5 A Current flows for time, t = 10⁻3 s Average back emf, e = dφ/dt---------------------Eq-1 Where, dφ = Change in flux = NAB ------------Read more
Length of the solenoid, 1 = 30 cm = 0.3 m
Area of cross-section, A = 25 cm2 = 25 x 10-4 m2
Number of turns on the solenoid, N = 500
Current in the solenoid, I = 2.5 A
Current flows for time, t = 10⁻3 s
Average back emf, e = dφ/dt———————Eq-1
Where, dφ = Change in flux = NAB ——————-Eq-2
B = Magnetic field strength =μ0NI/l——————–Eq-3
Where, μ0= Permeability of free space = 4π x 10-7 T m A-1
Using equations (2) and (3) in equation (1), we get
e = μ0N²IA/lt
= [(4π x 10-7) x (500)² x 2.5 x 25 x 10-4]/(0.3 x 10⁻3)
=6.5 V
Hence, the average back emf induced in the solenoid is 6.5 V.
See lessFigure 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?
Length of the rod, l = 15 cm = 0.15 m Magnetic field strength, B = 0.50 T Resistance of the closed loop, R = 9 mΩ = 9 x 10⁻³ Ω Ans (a). 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 InducedRead more
Length of the rod, l = 15 cm = 0.15 m
Magnetic field strength, B = 0.50 T
Resistance of the closed loop, R = 9 mΩ = 9 x 10⁻³ Ω
Ans (a).
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 given as:
e = Bvl
= 0.5 x 0.12 x 0.15 = 9 x 10-3 v = 9 mV
The polarity of the induced emf is such that end P shows positive while end Q shows negative ends.
Ans (b).
Yes; when key K is closed, excess charge is maintained by the continuous flow of current.
When key K is open, there is excess charge built up at both ends of the rods.
When key K is closed, excess charge is maintained by the continuous flow of current.
Ans (c).
Magnetic force is cancelled by the electric force set-up due to the excess charge of opposite nature at both ends of the rod.
There is no net force on the electrons in rod PQ when key K is open and the rod is moving uniformly. This is because magnetic force is cancelled by the electric force set-up due to the excess charge of opposite nature at both ends of the rods.
Ans (d).
Retarding force exerted on the rod, F = IBl Where,
I = Current flowing through the rod
= e/R = (9 x 10-3 )/(9 x 10-3 ) =1 A
Therefore, F = 1 x 0.5 x 0.15 =75 x 10-3 N
Ans (e).
9 mW; no power is expended when key K is open.
Speed of the rod, v = 12 cm/s = 0.12 m/s
Hence, power is given as:
P = Fv
= 75 x 10-3 x0.12 = 9×10-3 W = 9 mW
When key K is open, no power is expended.
Ans (f).
9 mW; power is provided by an external agent.
Power dissipated as heat = I2 R = (l)2 x 9x 10-3 = 9 mW.
The source of this power is an external agent.
Ans (g).
Zero
In this case, no emf is induced in the coil because the motion of the rod does not cut across the field lines.
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