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Page No 201: - Chapter 5 Magnetism & Matter Exercise Solutions class 12 ncert solutions Physics - SaraNextGen [2024]


Question 5.5:

A 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?

Answer:

Number of turns in the solenoid, = 800

Area of cross-section, A = 2.5 × 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 × 3 × 2.5 × 10−4

= 0.6 J T−1

Question 5.6:

If 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?

Answer:

Magnetic field strength, B = 0.25 T

Magnetic moment, M = 0.6 T−1

The angle θ, 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:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7027/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_m3a2a5239.gif

Question 5.7:

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)?

Answer:

(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, θ1 = 0°

Final angle between the axis and the magnetic field, θ2 = 90°

The work required to make the magnetic moment normal to the direction of magnetic field is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7028/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_m4098b48d.gif

(ii) Initial angle between the axis and the magnetic field, θ1 = 0°

Final angle between the axis and the magnetic field, θ2 = 180°

The work required to make the magnetic moment opposite to the direction of magnetic field is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7028/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_m2f338421.gif

(b)For case (i): https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7028/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_f084871.gif

∴Torque, https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7028/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_m377176a4.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7028/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_m3dc72375.gif

For case (ii): https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7028/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_mc976a24.gif

∴Torque, https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7028/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_m377176a4.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7028/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_34cfef39.gif

Question 5.8:

A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10−4 m2, carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane.

(a) What is the magnetic moment associated with the solenoid?

(b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 × 10−2 T is set up at an angle of 30º with the axis of the solenoid?

Answer:

Number of turns on the solenoid, n = 2000

Area of cross-section of the solenoid, A = 1.6 × 10−4 m2

Current in the solenoid, I = 4 A

(a)The magnetic moment along the axis of the solenoid is calculated as:

M = nAI

= 2000 × 1.6 × 10−4 × 4

= 1.28 Am2

(b)Magnetic field, B = 7.5 × 10−2 T

Angle between the magnetic field and the axis of the solenoid, θ = 30°

Torque, https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7030/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_m377176a4.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7030/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_65140acf.gif

Since the magnetic field is uniform, the force on the solenoid is zero. The torque on the solenoid is https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7030/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_m1871c0b7.gif

Question 5.9:

A circular coil of 16 turns and radius 10 cm carrying a current of 0.75 A rests with its plane normal to an external field of magnitude 5.0 × 10−2 T. The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of 2.0 s−1. What is the moment of inertia of the coil about its axis of rotation?

Answer:

Number of turns in the circular coil, N = 16

Radius of the coil, r = 10 cm = 0.1 m

Cross-section of the coil, A = πr2 = π × (0.1)2 m2

Current in the coil, I = 0.75 A

Magnetic field strength, B = 5.0 × 10−2 T

Frequency of oscillations of the coil, v = 2.0 s−1

∴Magnetic moment, M = NIAhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7031/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_m18e2a667.gif

= 16 × 0.75 × π × (0.1)2

= 0.377 J T−1

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7031/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_m1420f219.gif

Where,

I = Moment of inertia of the coil

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7031/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_m2f13ed74.gif

Hence, the moment of inertia of the coil about its axis of rotation is https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7031/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_29e3fa91.gif

Question 5.10:

A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22º with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.

Answer:

Horizontal component of earth’s magnetic field, BH = 0.35 G

Angle made by the needle with the horizontal plane = Angle of dip = https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7032/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_2f932a65.gif

Earth’s magnetic field strength = B

We can relate and BHas:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7032/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_2d2a8a73.gif

Hence, the strength of earth’s magnetic field at the given location is 0.377 G.

Question 5.11:

At a certain location in Africa, a compass points 12º west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points 60º above the horizontal. The horizontal component of the earth’s field is measured to be 0.16 G. Specify the direction and magnitude of the earth’s field at the location.

Answer:

Angle of declination,θ = 12°

Angle of dip, https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7034/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_51f97dd3.gif

Horizontal component of earth’s magnetic field, BH = 0.16 G

Earth’s magnetic field at the given location = B

We can relate and BHas:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7034/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_3ab1466b.gif

Earth’s magnetic field lies in the vertical plane, 12° West of the geographic meridian, making an angle of 60° (upward) with the horizontal direction. Its magnitude is 0.32 G.

Question 5.12:

A short bar magnet has a magnetic moment of 0.48 J T−1. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.

Answer:

Magnetic moment of the bar magnet, M = 0.48 J T−1

(a) Distance, d = 10 cm = 0.1 m

The magnetic field at distance d, from the centre of the magnet on the axis is given by the relation:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7035/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_3ac2a291.gif

Where,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7035/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_m144f80ba.gif = Permeability of free space = https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7035/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_47508db6.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7035/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_m3d7de6e9.gif

The magnetic field is along the S − N direction.

(b) The magnetic field at a distance of 10 cm (i.e., d = 0.1 m) on the equatorial line of the magnet is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7035/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_m553fc298.gif

The magnetic field is along the N − S direction.

Question 5.13:

A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. The earth’s magnetic field at the place is 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null−point (i.e., 14 cm) from the centre of the magnet? (At null points, field due to a magnet is equal and opposite to the horizontal component of earth’s magnetic field.)

Answer:

Earth’s magnetic field at the given place, H = 0.36 G

The magnetic field at a distance d, on the axis of the magnet is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7036/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_4bae188a.gif

Where,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7036/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_m144f80ba.gif= Permeability of free space

M = Magnetic moment

The magnetic field at the same distance d, on the equatorial line of the magnet is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7036/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_m1995a647.gif

Total magnetic field,https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7036/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_m29aaeebb.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7036/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_m15aebae1.gif

Hence, the magnetic field is 0.54 G in the direction of earth’s magnetic field.

Question 5.14:

If the bar magnet in exercise 5.13 is turned around by 180º, where will the new null points be located?

Answer:

The magnetic field on the axis of the magnet at a distance d1 = 14 cm, can be written as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7038/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_m4f1cc7e0.gif

Where,

= Magnetic moment

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7038/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_m144f80ba.gif= Permeability of free space

H = Horizontal component of the magnetic field at d1

If the bar magnet is turned through 180°, then the neutral point will lie on the equatorial line.

Hence, the magnetic field at a distance d2, on the equatorial line of the magnet can be written as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7038/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_6b58e996.gif

Equating equations (1) and (2), we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7038/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_m1a14a7da.gif

The new null points will be located 11.1 cm on the normal bisector.

Also Read : Page-No-202:-Chapter-5-Magnetism-&-Matter-Exercise-Solutions-class-12-ncert-solutions-Physics

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