Ans (a). Magnetic moment, M = 1.5 J T-1 , Magnetic field strength, B = 0.22 T (i) Initial angle between the axis and the magnetic field, 0₁ = 0° Final angle between the axis and the magnetic field, 02 = 90° The work required to make the magnetic moment normal to the direction of magnetic field is giRead more
Ans (a).
Magnetic moment, M = 1.5 J T-1 ,
Magnetic field strength, B = 0.22 T
(i) Initial angle between the axis and the magnetic field, 0₁ = 0°
Final angle between the axis and the magnetic field, 02 = 90°
The work required to make the magnetic moment normal to the direction of magnetic field is given as:
W = -MB(cos02 – cos 0₁)
= -1.5 x 0.22(cos 90° – cos 0°)
= -0.33(0-1)
= 0.33 J
(ii) Initial angle between the axis and the magnetic field, 0₁= 0°
Final angle between the axis and the magnetic field, 02 = 180°
The work required to make the magnetic moment opposite to the direction of magnetic field is given as:
Magnetic field strength, B = 0.25 T Magnetic moment, M = 0.6 T⁻¹ The angle 0, between the axis of the solenoid and the direction of the applied field is 30°. Therefore, the torque acting on the solenoid is given as: τ = MB sin O = 0.6 x 0.25 sin 30° = 7.5 x 10⁻2 J
Magnetic field strength, B = 0.25 T
Magnetic moment, M = 0.6 T⁻¹
The angle 0, between the axis of the solenoid and the direction of the applied field is 30°.
Therefore, the torque acting on the solenoid is given as:
Number of turns in the solenoid, n = 800, Area of cross-section, A = 2.5 x 10-4 m2 Current in the solenoid, I = 3.0 A A current-carrying solenoid behaves as a bar magnet because a magnetic field develops along its axis, i.e., along its length. The magnetic moment associated with the given current-caRead more
Number of turns in the solenoid, n = 800,
Area of cross-section, A = 2.5 x 10-4 m2
Current in the solenoid, I = 3.0 A
A current-carrying solenoid behaves as a bar magnet because a magnetic field develops along its axis, i.e., along its length.
The magnetic moment associated with the given current-carrying solenoid is calculated as:
Moment of the bar magnet, M = 0.32 J T-1 External magnetic field, B = 0.15 T Ans (a). The bar magnet is aligned along the magnetic field. This system is considered as being in stable equilibrium. Hence, the angle 0, between the bar magnet and the magnetic field is 0°. Potential energy of the systemRead more
Moment of the bar magnet, M = 0.32 J T-1
External magnetic field, B = 0.15 T
Ans (a).
The bar magnet is aligned along the magnetic field. This system is considered as being in stable equilibrium. Hence, the angle 0, between the bar magnet and the magnetic field is 0°.
Potential energy of the system = _MBos0 = -0.32 x 0.15 cos 0°
= -4.8 x 10⁻² J
Ans (b).
The bar magnet is oriented 180° to the magnetic field. Hence, it is in unstable equilibrium. 0 = 180° Potential energy = – MB cos 0 = -0.32×0.15 cos 180°
Magnetic field strength, B = 0.25 T Torque on the bar magnet, T = 4.5 x 10-2 J Angle between the bar magnet and the external magnetic field, 0 = 30° Torque is related to magnetic moment (M) as: T = MB sin 0 Therefore , M = T/Bsin 0 = (4.5 x 10-2 )/(0.25 x sin 30º) =0.36 J T⁻¹ Hence, the magnRead more
Magnetic field strength, B = 0.25 T
Torque on the bar magnet, T = 4.5 x 10-2 J
Angle between the bar magnet and the external magnetic field, 0 = 30° Torque is related to magnetic moment (M) as: T = MB sin 0
Therefore , M = T/Bsin 0
= (4.5 x 10-2 )/(0.25 x sin 30º) =0.36 J T⁻¹
Hence, the magnetic moment of the magnet is 0.36 IT-1.
A bar magnet of magnetic moment 1.5 J T⁻¹ lies aligned with the direction of a uniform magnetic field of 0.22 T. (a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction? (b) What is the torque on the magnet in cases (i) and (ii)?
Ans (a). Magnetic moment, M = 1.5 J T-1 , Magnetic field strength, B = 0.22 T (i) Initial angle between the axis and the magnetic field, 0₁ = 0° Final angle between the axis and the magnetic field, 02 = 90° The work required to make the magnetic moment normal to the direction of magnetic field is giRead more
Ans (a).
Magnetic moment, M = 1.5 J T-1 ,
Magnetic field strength, B = 0.22 T
(i) Initial angle between the axis and the magnetic field, 0₁ = 0°
Final angle between the axis and the magnetic field, 02 = 90°
The work required to make the magnetic moment normal to the direction of magnetic field is given as:
W = -MB(cos02 – cos 0₁)
= -1.5 x 0.22(cos 90° – cos 0°)
= -0.33(0-1)
= 0.33 J
(ii) Initial angle between the axis and the magnetic field, 0₁= 0°
Final angle between the axis and the magnetic field, 02 = 180°
The work required to make the magnetic moment opposite to the direction of magnetic field is given as:
W = -MB(cos02 – cos 0₁)
= -1.5 x 0.22(cos 180 – cos 0°)
= —0.33(—1 -1)
= 0.66 J
Ans (b).
For case (i): 0 = 02= 900
Therefore ,Torque, τ = MBsin 6
= 1.5×0.22 sin 90°
= 0.33 J
For case (ii): 0 = 02= 180°
Therefore Torque, τ = MB sin 0
= MBsin 180° = 0 J
See lessIf the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30° with the direction of applied field?
Magnetic field strength, B = 0.25 T Magnetic moment, M = 0.6 T⁻¹ The angle 0, between the axis of the solenoid and the direction of the applied field is 30°. Therefore, the torque acting on the solenoid is given as: τ = MB sin O = 0.6 x 0.25 sin 30° = 7.5 x 10⁻2 J
Magnetic field strength, B = 0.25 T
Magnetic moment, M = 0.6 T⁻¹
The angle 0, between the axis of the solenoid and the direction of the applied field is 30°.
Therefore, the torque acting on the solenoid is given as:
τ = MB sin O
= 0.6 x 0.25 sin 30°
= 7.5 x 10⁻2 J
See lessA closely wound solenoid of 800 turns and area of cross section 2.5 × 10⁻⁴ m² carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?
Number of turns in the solenoid, n = 800, Area of cross-section, A = 2.5 x 10-4 m2 Current in the solenoid, I = 3.0 A A current-carrying solenoid behaves as a bar magnet because a magnetic field develops along its axis, i.e., along its length. The magnetic moment associated with the given current-caRead more
Number of turns in the solenoid, n = 800,
Area of cross-section, A = 2.5 x 10-4 m2
Current in the solenoid, I = 3.0 A
A current-carrying solenoid behaves as a bar magnet because a magnetic field develops along its axis, i.e., along its length.
The magnetic moment associated with the given current-carrying solenoid is calculated as:
M = n I A = 800 x 3 x 2.5 x 10-4 = 0.6 J T–1
See lessA short bar magnet of magnetic moment m = 0.32 JT⁻¹ is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?
Moment of the bar magnet, M = 0.32 J T-1 External magnetic field, B = 0.15 T Ans (a). The bar magnet is aligned along the magnetic field. This system is considered as being in stable equilibrium. Hence, the angle 0, between the bar magnet and the magnetic field is 0°. Potential energy of the systemRead more
Moment of the bar magnet, M = 0.32 J T-1
External magnetic field, B = 0.15 T
Ans (a).
The bar magnet is aligned along the magnetic field. This system is considered as being in stable equilibrium. Hence, the angle 0, between the bar magnet and the magnetic field is 0°.
Potential energy of the system = _MBos0 = -0.32 x 0.15 cos 0°
= -4.8 x 10⁻² J
Ans (b).
The bar magnet is oriented 180° to the magnetic field. Hence, it is in unstable equilibrium. 0 = 180° Potential energy = – MB cos 0 = -0.32×0.15 cos 180°
= 4.8 x 10⁻² J
See lessA short bar magnet placed with its axis at 30º with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 × 10⁻² J. What is the magnitude of magnetic moment of the magnet?
Magnetic field strength, B = 0.25 T Torque on the bar magnet, T = 4.5 x 10-2 J Angle between the bar magnet and the external magnetic field, 0 = 30° Torque is related to magnetic moment (M) as: T = MB sin 0 Therefore , M = T/Bsin 0 = (4.5 x 10-2 )/(0.25 x sin 30º) =0.36 J T⁻¹ Hence, the magnRead more
Magnetic field strength, B = 0.25 T
Torque on the bar magnet, T = 4.5 x 10-2 J
Angle between the bar magnet and the external magnetic field, 0 = 30° Torque is related to magnetic moment (M) as: T = MB sin 0
Therefore , M = T/Bsin 0
= (4.5 x 10-2 )/(0.25 x sin 30º) =0.36 J T⁻¹
Hence, the magnetic moment of the magnet is 0.36 IT-1.
See less