Area of the small flat search coil, A = 2 cm2= 2 x 10-4 m2 Number of turns on the coil, N = 25 Total charge flowing in the coil, Q = 7.5 mC = 7.5 x 10-3 C Total resistance of the coil and galvanometer, R = 0.50Ω Induced current in the coil, I = [Induced emf (e)]/R------------------------Eq-1 InducedRead more
Area of the small flat search coil, A = 2 cm2= 2 x 10-4 m2
Number of turns on the coil, N = 25
Total charge flowing in the coil, Q = 7.5 mC = 7.5 x 10-3 C
Total resistance of the coil and galvanometer, R = 0.50Ω
Induced current in the coil, I = [Induced emf (e)]/R————————Eq-1
Induced emf is given as: e = —N dφ/dt……………………Eq-2
Where, dφ = Charge in flux
Combining equations (1) and (2), we get
I = – (N dφ/dt)/R
=> I dt = -Ndφ/R —————————Eq-3
Initial flux through the coil,φi = BA
Where, B = Magnetic field strength
Final flux through the coil, φf = 0
Integrating equation (3) on both sides, we have
⌠Idt = -N/R Φi⌠Φf dΦ { NOTE- “⌠” is sign of integration}
But total charge ,Q = ⌠I dt.
Therefore, Q = -N/R (Φf-Φi) = -N/R (-Φi)= + NΦi/R
=> Q = NBA/R
Therefore,B = QR /NA
= (7.5 x 10-3 x 0.5)/(25 x 2 x 10-4)
= 0.75 T
Hence, the field strength of the magnet is 0.75 T.
Side of the square loop, s = 12 cm = 0.12 m Area of the square loop, A = 0.12 x 0.12 = 0.0144 m2 Velocity of the loop, v = 8 cm/s = 0.08 m/s Gradient of the magnetic field along negative x-direction, dB/dx = 10⁻³ Tcm⁻¹= 10⁻¹ T m⁻¹ And, rate of decrease of the magnetic field, dB/dt = 10⁻³ Ts⁻¹ ResistRead more
Side of the square loop, s = 12 cm = 0.12 m Area of the square loop, A = 0.12 x 0.12 = 0.0144 m2 Velocity of the loop, v = 8 cm/s = 0.08 m/s Gradient of the magnetic field along negative x-direction,
dB/dx = 10⁻³ Tcm⁻¹= 10⁻¹ T m⁻¹ And, rate of decrease of the magnetic field,
dB/dt = 10⁻³ Ts⁻¹
Resistance of the loop, R = 4.5 mΩ = 4.5 x 10-3 Ω
Rate of change of the magnetic flux due to the motion of the loop in a non-uniform magnetic field is given as:
dφ/dt =A x dB/dx x v
= 144 x 10⁻⁴m² x 10⁻¹ x 0.08
=11.52 x 10⁻⁵ Tm² s⁻¹
Rate of change of the flux due to explicit time variation in field B is given as:
dφ´/dt =A x dB/dt
= 144 x 10⁻⁴ x 10⁻³
= 1.44 x 10⁻⁵ Tm² s⁻¹
Since the rate of change of the flux is the induced emf, the total induced emf in the loop can be calculated as:
e = 1.44 x 10⁻⁵+ 11.52 x 10⁻⁵
= 12.96 x 10⁻⁵ V
Therefore, Induced current, i = e/R
= (12.96 x 10⁻⁵)/(4.5 x 10-3) = 2.88 x 10⁻² A
Hence, the direction of the induced current is such that there is an increase in the flux through the loop along positive z-direction.
Sides of the rectangular loop are 8 cm and 2 cm. Hence, area of the rectangular wire loop, A = length x width = 8 x 2 = 16 cm2 = 16 x 10⁻⁴ m2 Initial value of the magnetic field, B' = 0.3 T Rate of decrease of the magnetic field, dB/dt = 0.02 T/s Emf developed in the loop is given as: e = dφ/dt WherRead more
Sides of the rectangular loop are 8 cm and 2 cm.
Hence, area of the rectangular wire loop,
A = length x width = 8 x 2 = 16 cm2 = 16 x 10⁻⁴ m2
Initial value of the magnetic field, B’ = 0.3 T
Rate of decrease of the magnetic field, dB/dt = 0.02 T/s
Emf developed in the loop is given as: e = dφ/dt
Where, dφ= Change in flux through the loop area = AB
Therefore, e = d(AB)/dt = AdB/dt
= 16 x 10⁻⁴ x 0.02 = 0.32 x 10⁻⁴ V
Resistance of the loop, R = 1.6 Ω
The current induced in the loop is given as: i = e/R
= (0.32 x 10⁻⁴)/1.6 = 2 x 10⁻⁵ A
Power dissipated in the loop in the form of heat is given as:
P = i2R
= (2 x 10⁻5)2 x 1.6 = 6.4 x 10⁻¹⁰ W The source of this heat loss is an external agent, which is responsible for changing the magnetic field with time.
Speed of the jet plane, v = 1800 km/h = 500 m/s Wing span of jet plane, l = 25 m Earth’s magnetic field strength, B = 5.0 x 10-4 T Angle of dip, δ = 30° Vertical component of Earth's magnetic field, Bv = B sin δ = 5 x 10_4sin 30° = 2.5 x 10-4 T Voltage difference between the ends of the wing can beRead more
Speed of the jet plane, v = 1800 km/h = 500 m/s
Wing span of jet plane, l = 25 m
Earth’s magnetic field strength, B = 5.0 x 10-4 T
Angle of dip, δ = 30°
Vertical component of Earth’s magnetic field,
Bv = B sin δ = 5 x 10_4sin 30°
= 2.5 x 10-4 T
Voltage difference between the ends of the wing can be calculated as: e = (Bv) x l x v
= 2.5 x 10⁻⁴x 25 x 500 = 3.125 V
Hence, the voltage difference developed between the ends of the wings is 3.125 V.
Mutual inductance of a pair of coils, μ = 1.5 H Initial current, I₁= 0 A Final current I₂ = 20 A Change in current, dl = I2 — I₁= 20 — 0 = 20 A Time taken for the change, t = 0.5 s Induced emf e = dφ/dt ------------------Eq-1 Where, dφ is the change in the flux linkage with the coil. Emf is relatedRead more
Mutual inductance of a pair of coils, μ = 1.5 H
Initial current, I₁= 0 A Final current I₂ = 20 A
Change in current, dl = I2 — I₁= 20 — 0 = 20 A
Time taken for the change, t = 0.5 s
Induced emf e = dφ/dt ——————Eq-1
Where, dφ is the change in the flux linkage with the coil.
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.
Area of the small flat search coil, A = 2 cm2= 2 x 10-4 m2 Number of turns on the coil, N = 25 Total charge flowing in the coil, Q = 7.5 mC = 7.5 x 10-3 C Total resistance of the coil and galvanometer, R = 0.50Ω Induced current in the coil, I = [Induced emf (e)]/R------------------------Eq-1 InducedRead more
Area of the small flat search coil, A = 2 cm2= 2 x 10-4 m2
Number of turns on the coil, N = 25
Total charge flowing in the coil, Q = 7.5 mC = 7.5 x 10-3 C
Total resistance of the coil and galvanometer, R = 0.50Ω
Induced current in the coil, I = [Induced emf (e)]/R————————Eq-1
Induced emf is given as: e = —N dφ/dt……………………Eq-2
Where, dφ = Charge in flux
Combining equations (1) and (2), we get
I = – (N dφ/dt)/R
=> I dt = -Ndφ/R —————————Eq-3
Initial flux through the coil,φi = BA
Where, B = Magnetic field strength
Final flux through the coil, φf = 0
Integrating equation (3) on both sides, we have
⌠Idt = -N/R Φi⌠Φf dΦ { NOTE- “⌠” is sign of integration}
But total charge ,Q = ⌠I dt.
Therefore, Q = -N/R (Φf-Φi) = -N/R (-Φi)= + NΦi/R
=> Q = NBA/R
Therefore,B = QR /NA
= (7.5 x 10-3 x 0.5)/(25 x 2 x 10-4)
= 0.75 T
Hence, the field strength of the magnet is 0.75 T.
See lessA 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 10⁻³T cm⁻¹ along the negative x-direction (that is it increases by 10⁻³T cm⁻¹ as one moves in the negative x-direction), and it is decreasing in time at the rate of 10⁻³T s⁻¹. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 mΩ..
Side of the square loop, s = 12 cm = 0.12 m Area of the square loop, A = 0.12 x 0.12 = 0.0144 m2 Velocity of the loop, v = 8 cm/s = 0.08 m/s Gradient of the magnetic field along negative x-direction, dB/dx = 10⁻³ Tcm⁻¹= 10⁻¹ T m⁻¹ And, rate of decrease of the magnetic field, dB/dt = 10⁻³ Ts⁻¹ ResistRead more
Side of the square loop, s = 12 cm = 0.12 m
Area of the square loop, A = 0.12 x 0.12 = 0.0144 m2
Velocity of the loop, v = 8 cm/s = 0.08 m/s
Gradient of the magnetic field along negative x-direction,
dB/dx = 10⁻³ Tcm⁻¹= 10⁻¹ T m⁻¹
And, rate of decrease of the magnetic field,
dB/dt = 10⁻³ Ts⁻¹
Resistance of the loop, R = 4.5 mΩ = 4.5 x 10-3 Ω
Rate of change of the magnetic flux due to the motion of the loop in a non-uniform magnetic field is given as:
dφ/dt =A x dB/dx x v
= 144 x 10⁻⁴m² x 10⁻¹ x 0.08
=11.52 x 10⁻⁵ Tm² s⁻¹
Rate of change of the flux due to explicit time variation in field B is given as:
dφ´/dt =A x dB/dt
= 144 x 10⁻⁴ x 10⁻³
= 1.44 x 10⁻⁵ Tm² s⁻¹
Since the rate of change of the flux is the induced emf, the total induced emf in the loop can be calculated as:
e = 1.44 x 10⁻⁵+ 11.52 x 10⁻⁵
= 12.96 x 10⁻⁵ V
Therefore, Induced current, i = e/R
= (12.96 x 10⁻⁵)/(4.5 x 10-3) = 2.88 x 10⁻² A
Hence, the direction of the induced current is such that there is an increase in the flux through the loop along positive z-direction.
See lessSuppose 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 1.6Ω how much power is dissipated by the loop as heat? What is the source of this power?
Sides of the rectangular loop are 8 cm and 2 cm. Hence, area of the rectangular wire loop, A = length x width = 8 x 2 = 16 cm2 = 16 x 10⁻⁴ m2 Initial value of the magnetic field, B' = 0.3 T Rate of decrease of the magnetic field, dB/dt = 0.02 T/s Emf developed in the loop is given as: e = dφ/dt WherRead more
Sides of the rectangular loop are 8 cm and 2 cm.
Hence, area of the rectangular wire loop,
A = length x width = 8 x 2 = 16 cm2 = 16 x 10⁻⁴ m2
Initial value of the magnetic field, B’ = 0.3 T
Rate of decrease of the magnetic field, dB/dt = 0.02 T/s
Emf developed in the loop is given as: e = dφ/dt
Where, dφ= Change in flux through the loop area = AB
Therefore, e = d(AB)/dt = AdB/dt
= 16 x 10⁻⁴ x 0.02 = 0.32 x 10⁻⁴ V
Resistance of the loop, R = 1.6 Ω
The current induced in the loop is given as: i = e/R
= (0.32 x 10⁻⁴)/1.6 = 2 x 10⁻⁵ A
Power dissipated in the loop in the form of heat is given as:
P = i2R
= (2 x 10⁻5)2 x 1.6 = 6.4 x 10⁻¹⁰ W
See lessThe source of this heat loss is an external agent, which is responsible for changing the magnetic field with time.
A jet plane is travelling towards 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 x 10⁻⁴ T and the dip angle is 30°.
Speed of the jet plane, v = 1800 km/h = 500 m/s Wing span of jet plane, l = 25 m Earth’s magnetic field strength, B = 5.0 x 10-4 T Angle of dip, δ = 30° Vertical component of Earth's magnetic field, Bv = B sin δ = 5 x 10_4sin 30° = 2.5 x 10-4 T Voltage difference between the ends of the wing can beRead more
Speed of the jet plane, v = 1800 km/h = 500 m/s
Wing span of jet plane, l = 25 m
Earth’s magnetic field strength, B = 5.0 x 10-4 T
Angle of dip, δ = 30°
Vertical component of Earth’s magnetic field,
Bv = B sin δ = 5 x 10_4sin 30°
= 2.5 x 10-4 T
Voltage difference between the ends of the wing can be calculated as: e = (Bv) x l x v
= 2.5 x 10⁻⁴x 25 x 500 = 3.125 V
Hence, the voltage difference developed between the ends of the wings is 3.125 V.
See lessA 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?
Mutual inductance of a pair of coils, μ = 1.5 H Initial current, I₁= 0 A Final current I₂ = 20 A Change in current, dl = I2 — I₁= 20 — 0 = 20 A Time taken for the change, t = 0.5 s Induced emf e = dφ/dt ------------------Eq-1 Where, dφ is the change in the flux linkage with the coil. Emf is relatedRead more
Mutual inductance of a pair of coils, μ = 1.5 H
Initial current, I₁= 0 A Final current I₂ = 20 A
Change in current, dl = I2 — I₁= 20 — 0 = 20 A
Time taken for the change, t = 0.5 s
Induced emf e = dφ/dt ——————Eq-1
Where, dφ is the change in the flux linkage with the coil.
Emf is related with mutual inductance as:
e = μ dI /dt —————————-Eq-2
Equating equations (1) and (2), we get
dφ/dt = μ dI /dt
dφ = 1.5x(20)
= 30 Wb
Hence, the change in the flux linkage is 30 Wb.
See less