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 x 10⁻4 m2 Current carried by the solenoid changes from 2 A to 4 A. Therefore, change in current in the solenoid, di = 4- 2 = 2A Change iRead more
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 x 10⁻4 m2
Current carried by the solenoid changes from 2 A to 4 A.
Therefore, change in current in the solenoid, di = 4- 2 = 2A
Change in time, dt = 0.1 s
Induced emf in the solenoid is given by Faraday’s law as: e =dφ/dt ————Eq-1
Where, φ= Induced flux through the small loop =BA————–Eq -2 B = Magnetic field =μ0ni————–Eq -3
μ0 = Permeability of free space = 4π x10-7 H/m
Hence, equation (i) reduces to:
e =d/dt (BA)
= Aμ0n (di/dt)
= 2 x 10⁻4 x 4π x10-7 x 1500 x 2/ 0.1
= 7.54 x 10⁻⁶ V
Hence, the induced voltage in the loop is 7.54 x 10-6V
Ans (a). As the loop changes from irregular to circular shape, its area increases. Hence, the magnetic flux linked with it increases. According to Lenz's law, the induced current should produce magnetic flux in the opposite direction of original flux. For this induced current should flow in the antiRead more
Ans (a).
As the loop changes from irregular to circular shape, its area increases. Hence, the magnetic
flux linked with it increases. According to Lenz’s law, the induced current should produce magnetic flux in the opposite direction of original flux. For this induced current should flow in the anti-clock wise direction. It means the direction of current will be along adcba.
Ans (b).
As the circular loop is being deformed into a narrow straight wire, its area decreases. The magnetic field linked with it also decreases. By Lenz’s law, the induced current should produce a flux in the direction of original flux. For this, the induced current should flow in the anti-clock wise direction. It means the direction of current will be along a’d’c’b’.
The direction of the induced current in a closed loop is given by Lenz's law, it states that: The polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it. The given pairs of figures show the direction of the induced current when tRead more
The direction of the induced current in a closed loop is given by Lenz’s law, it states that: The polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it.
The given pairs of figures show the direction of the induced current when the North pole of a bar magnet is moved towards and away from a closed loop respectively.
Using Lenz’s rule, the direction of the induced current in the given situations can be predicted as follows:
Ans (a).The direction of the induced current is along qrpq.
Ans (b).The direction of the induced current is along prqp.
Ans (c).The direction of the induced current is along yzxy.
Ans (d).The direction of the induced current is along zyxz.
Ans (e).The direction of the induced current is along xryx.
Ans (f).No current is induced since the field lines are lying in the plane of the closed loop.
Out of the two relations given, only one is in accordance with classical physics. The magnetic momemtum vector which as a result of orbital Angular Moment is given by , μl=−e/2m l It follows from the definitions ofμl and l. μl=iA=(−e/T)πr2 ...(i) Angular momentum, l=mvr=m(2πr/T)r ...(iiRead more
Out of the two relations given, only one is in accordance with classical physics.
The magnetic momemtum vector which as a result of orbital Angular Moment is given by ,
μl=−e/2m l
It follows from the definitions ofμl and l.
μl=iA=(−e/T)πr2 …(i)
Angular momentum, l=mvr=m(2πr/T)r …(ii)
where r is the radius of the circular orbit, which the electron of mass m and charge (−e) completes in time T.
Divide (i) by (ii), μl/l=[(−e/T)πr2 ]/(m(2πr/T)r )=−e/2m
∴μl=(−e/2m)l
Clearly μl and l will be antiparallel (both being normal to the plane of the orbit)
In contrast, μs/S=e/m. It is obtained on the basis of quantum mechanics.
Mean radius of a Rowland ring, r = 15 cm = 0.15 m Number of turns on a ferromagnetic core, N = 3500 Relative permeability of the core material, μ =800 Magnetising current, I = 1.2 A The magnetic field is given by the relation: B = μrμ0 IN/2πr Where, [μ0 = Permeability of free space = 4π x 10-7 T m ARead more
Mean radius of a Rowland ring, r = 15 cm = 0.15 m
Number of turns on a ferromagnetic core, N = 3500
Relative permeability of the core material, μ =800
Magnetising current, I = 1.2 A
The magnetic field is given by the relation:
B = μrμ0 IN/2πr
Where, [μ0 = Permeability of free space = 4π x 10-7 T m A-1
B = (800 x 4π x 10-7 x 1.2 x 3500) / (2πx 0.15) =4.48 T
Therefore, the magnetic field in the core is 4.48 T.
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?
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 x 10⁻4 m2 Current carried by the solenoid changes from 2 A to 4 A. Therefore, change in current in the solenoid, di = 4- 2 = 2A Change iRead more
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 x 10⁻4 m2
Current carried by the solenoid changes from 2 A to 4 A.
Therefore, change in current in the solenoid, di = 4- 2 = 2A
Change in time, dt = 0.1 s
Induced emf in the solenoid is given by Faraday’s law as: e =dφ/dt ————Eq-1
Where, φ= Induced flux through the small loop =BA————–Eq -2 B = Magnetic field =μ0ni————–Eq -3
μ0 = Permeability of free space = 4π x10-7 H/m
Hence, equation (i) reduces to:
e =d/dt (BA)
= Aμ0n (di/dt)
= 2 x 10⁻4 x 4π x10-7 x 1500 x 2/ 0.1
= 7.54 x 10⁻⁶ V
Hence, the induced voltage in the loop is 7.54 x 10-6V
See lessUse 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.
Ans (a). As the loop changes from irregular to circular shape, its area increases. Hence, the magnetic flux linked with it increases. According to Lenz's law, the induced current should produce magnetic flux in the opposite direction of original flux. For this induced current should flow in the antiRead more
Ans (a).
As the loop changes from irregular to circular shape, its area increases. Hence, the magnetic
flux linked with it increases. According to Lenz’s law, the induced current should produce magnetic flux in the opposite direction of original flux. For this induced current should flow in the anti-clock wise direction. It means the direction of current will be along adcba.
Ans (b).
As the circular loop is being deformed into a narrow straight wire, its area decreases. The magnetic field linked with it also decreases. By Lenz’s law, the induced current should produce a flux in the direction of original flux. For this, the induced current should flow in the anti-clock wise direction. It means the direction of current will be along a’d’c’b’.
See lessPredict the direction of induced current in the situations described by the following Figures. 6.18(a) to (f).
The direction of the induced current in a closed loop is given by Lenz's law, it states that: The polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it. The given pairs of figures show the direction of the induced current when tRead more
The direction of the induced current in a closed loop is given by Lenz’s law, it states that:
The polarity of induced emf is such that it tends to produce a current
which opposes the change in magnetic flux that produced it.
The given pairs of figures show the direction of the induced current when the North pole of a bar magnet is moved towards and away from a closed loop respectively.
Using Lenz’s rule, the direction of the induced current in the given situations can be predicted as follows:
Ans (a).The direction of the induced current is along qrpq.
Ans (b).The direction of the induced current is along prqp.
Ans (c).The direction of the induced current is along yzxy.
Ans (d).The direction of the induced current is along zyxz.
Ans (e).The direction of the induced current is along xryx.
Ans (f).No current is induced since the field lines are lying in the plane of the closed loop.
See lessThe magnetic moment vectors µs and µl associated with the intrinsic spin angular momentum S and orbital angular momentum l, respectively, of an electron are predicted by quantum theory (and verified experimentally to a high accuracy) to be given by: µs = –(e/m) S, µl = –(e/2m)l Which of these relations is in accordance with the result expected classically? Outline the derivation of the classical result.
Out of the two relations given, only one is in accordance with classical physics. The magnetic momemtum vector which as a result of orbital Angular Moment is given by , μl=−e/2m l It follows from the definitions ofμl and l. μl=iA=(−e/T)πr2 ...(i) Angular momentum, l=mvr=m(2πr/T)r ...(iiRead more
Out of the two relations given, only one is in accordance with classical physics.
The magnetic momemtum vector which as a result of orbital Angular Moment is given by ,
μl=−e/2m l
It follows from the definitions ofμl and l.
μl=iA=(−e/T)πr2 …(i)
Angular momentum, l=mvr=m(2πr/T)r …(ii)
where r is the radius of the circular orbit, which the electron of mass m and charge (−e) completes in time T.
Divide (i) by (ii), μl/l=[(−e/T)πr2 ]/(m(2πr/T)r )=−e/2m
∴μl=(−e/2m)l
Clearly μl and l will be antiparallel (both being normal to the plane of the orbit)
In contrast, μs/S=e/m. It is obtained on the basis of quantum mechanics.
See lessA Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core of relative permeability 800. What is the magnetic field B in the core for a magnetising current of 1.2 A?
Mean radius of a Rowland ring, r = 15 cm = 0.15 m Number of turns on a ferromagnetic core, N = 3500 Relative permeability of the core material, μ =800 Magnetising current, I = 1.2 A The magnetic field is given by the relation: B = μrμ0 IN/2πr Where, [μ0 = Permeability of free space = 4π x 10-7 T m ARead more
Mean radius of a Rowland ring, r = 15 cm = 0.15 m
Number of turns on a ferromagnetic core, N = 3500
Relative permeability of the core material, μ =800
Magnetising current, I = 1.2 A
The magnetic field is given by the relation:
B = μrμ0 IN/2πr
Where, [μ0 = Permeability of free space = 4π x 10-7 T m A-1
B = (800 x 4π x 10-7 x 1.2 x 3500) / (2πx 0.15) =4.48 T
Therefore, the magnetic field in the core is 4.48 T.
See less