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Page No 203: - Chapter 5 Magnetism & Matter Additional Exercise Solutions class 12 ncert solutions Physics - SaraNextGen [2024]


Question 5.20:

A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 45º with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east.

(a) Determine the horizontal component of the earth’s magnetic field at the location.

(b) The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of 90º in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.

Answer:

Number of turns in the circular coil, N = 30

Radius of the circular coil, r = 12 cm = 0.12 m

Current in the coil, I = 0.35 A

Angle of dip, δ = 45°

(a) The magnetic field due to current I, at a distance r, is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7044/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_mea1a7b2.gif

Where,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7044/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_m144f80ba.gif = Permeability of free space = 4π × 10−7 Tm A−1

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7044/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_m391f7a52.gif

= 5.49 × 10−5 T

The compass needle points from West to East. Hence, the horizontal component of earth’s magnetic field is given as:

BH = Bsin δ

= 5.49 × 10−5 sin 45° = 3.88 × 10−5 T = 0.388 G

(b) When the current in the coil is reversed and the coil is rotated about its vertical axis by an angle of 90 º, the needle will reverse its original direction. In this case, the needle will point from East to West.

Question 5.21:

A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is 60º, and one of the fields has a magnitude of 1.2 × 10−2 T. If the dipole comes to stable equilibrium at an angle of 15º with this field, what is the magnitude of the other field?

Answer:

Magnitude of one of the magnetic fields, B1 = 1.2 × 10−2 T

Magnitude of the other magnetic field = B2

Angle between the two fields, θ = 60°

At stable equilibrium, the angle between the dipole and field B1θ1 = 15°

Angle between the dipole and field B2θ2 = θ − θ= 60° − 15° = 45°

At rotational equilibrium, the torques between both the fields must balance each other.

∴Torque due to field B1 = Torque due to field B2

MB1 sinθ1 = MB2 sinθ2

Where,

M = Magnetic moment of the dipole

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7045/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_m1316971d.gif

Hence, the magnitude of the other magnetic field is 4.39 × 10−3 T.

Question 5.22:

A monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of 0.04 G normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30 cm (me= 9.11 × 10−19 C). [Note: Data in this exercise are so chosen that the Answer will give you an idea of the effect of earth’s magnetic field on the motion of the electron beam from the electron gun to the screen in a TV set.]

Answer:

Energy of an electron beam, E = 18 keV = 18 × 103 eV

Charge on an electron, e = 1.6 × 10−19 C

E = 18 × 103 × 1.6 × 10−19 J

Magnetic field, B = 0.04 G

Mass of an electron, me = 9.11 × 10−19 kg

Distance up to which the electron beam travels, = 30 cm = 0.3 m

We can write the kinetic energy of the electron beam as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7046/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_2d1bf835.gif

The electron beam deflects along a circular path of radius, r.

The force due to the magnetic field balances the centripetal force of the path.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7046/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_336d5e0a.gif

Let the up and down deflection of the electron beam be https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7046/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_5427dcd8.gif

Where,

θ = Angle of declination

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7046/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_m16422cca.gif

Therefore, the up and down deflection of the beam is 3.9 mm.

Question 5.23:

A sample of paramagnetic salt contains 2.0 × 1024 atomic dipoles each of dipole moment 1.5 × 10−23 J T−1. The sample is placed under a homogeneous magnetic field of 0.64 T, and cooled to a temperature of 4.2 K. The degree of magnetic saturation achieved is equal to 15%. What is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K? (Assume Curie’s law)

Answer:

Number of atomic dipoles, n = 2.0 × 1024

Dipole moment of each atomic dipole, M = 1.5 × 10−23 J T−1

When the magnetic field, B1 = 0.64 T

The sample is cooled to a temperature, T1 = 4.2°K

Total dipole moment of the atomic dipole, Mtot = n × M

= 2 × 1024 × 1.5 × 10−23

= 30 J T−1

Magnetic saturation is achieved at 15%.

Hence, effective dipole moment, https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7047/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_mc77fc28.gif

When the magnetic field, B2 = 0.98 T

Temperature, T2 = 2.8°K

Its total dipole moment = M2

According to Curie’s law, we have the ratio of two magnetic dipoles as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7047/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_m656844ad.gif

Therefore, https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7047/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_2321cba0.gif is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K.

Question 5.24:

A 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 in the core for a magnetising current of 1.2 A?

Answer:

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, https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7048/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_2e888057.gif

Magnetising current, I = 1.2 A

The magnetic field is given by the relation:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7048/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_m2f2f9189.gif

Where,

μ0 = Permeability of free space = 4π × 10−7 Tm A−1

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/249/7048/NS_17-11-08_Sravana_12_Physics_5_25_NRJ_LVN_html_m23ee5885.gif

Therefore, the magnetic field in the core is 4.48 T.

Question 5.25:

The magnetic moment vectors μsand μlassociated with the intrinsic spin angular momentum 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/mS,

μl = –(e/2m)l

Which of these relations is in accordance with the result expected classically? Outline the derivation of the classical result.

Answer:

The magnetic moment associated with the orbital angular momentum is valid with the classical mechanics.

The magnetic moment associated with the orbital angular momentum is given as

https://img-nm.mnimgs.com/img/study_content/editlive_ncert/89/2012_04_26_15_28_39/mathmlequation588578432798991390.png

For current and area of cross-section A, we have the relation:

Magnetic moment

https://img-nm.mnimgs.com/img/study_content/editlive_ncert/89/2012_04_26_15_28_39/mathmlequation3917595074262809681.png………………………….(1)

Where,

e= Charge of the electron

r= Radius of the circular orbit

T= Time taken to complete one rotation around the circular orbit of radius r

Orbital angular momentum, l= mvr

https://img-nm.mnimgs.com/img/study_content/editlive_ncert/89/2012_04_26_15_28_39/mathmlequation2285765333575880216.png…………………………….(2)

Where,

m= Mass of the electron

v= Velocity of the electron

r= Radius of the circular orbit

Dividing equation (1) by equation (2), we get:

https://img-nm.mnimgs.com/img/study_content/editlive_ncert/89/2012_04_26_15_28_39/mathmlequation2566931510131227133.png

Also Read : INTRODUCTION-Chapter-6-Electromagnetic-Induction-Exercise-Solutions-class-12-ncert-solutions-Physics

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