Thursday, January 13, 2022

H C Verma solutions, MAGNETIC PROPERTIES OF MATTER, Chapter-37, Exercises, Concepts of Physics, Part-II

Magnetic Properties of Matter


EXERCISES


    1. The magnetic intensity H at the center of a long solenoid carrying a current of 2.0 A, is found to be 1500 A/m. Find the number of turns per centimeter of the solenoid.    


ANSWER: The magnetic intensity at the center of a long solenoid, 

H =B/µₒ =µₒni/µₒ

→H =ni.  

Given i =2 A, H =1500 A/m. So,

1500 =n*2

→n =750 turns/m

     =750/100 turns/cm

     =7.5 turns/cm.          




 

    2. A rod is inserted as the core in the current-carrying solenoid of the previous problem, (a) What is the magnetic intensity H at the center? (b) If the magnetization I of the core is found to be 0.12 A/m, find the susceptibility of the material of the rod. (c) Is the material paramagnetic, diamagntic or ferromagnetic?      


ANSWER: (a) Since the rod is inserted as a core, it will be long as the solenoid. So at the center, the end effects may be neglected. Thus there will be no effect of the rod on the magnetic intensity at the center. The magnetic intensity H will remain the same i.e. equal to 1500 A/m. 


      (b) Given, magnetization of the core

 I = 0.12 A/m.

Since I =χH

(Where χ is susceptibility)

→χ =I/H

     = 0.12/1500

     = 8.0x10⁻⁵.


(c) Since the susceptibility of the material χ, is positive and very small it is paramagnetic. 

        



 

    3. The magnetic field inside a long solenoid having 50 turns/cm is increased from 2.5x10⁻³ T to 2.5 T when an iron core of cross-sectional area 4 cm² is inserted into it. Find (a) the current in the solenoid, (b) the magnetization I of the core, and (c) the pole strength developed in the core.     


ANSWER: (a) The magnetic field inside the solenoid, B =2.5x10⁻³ T. 

 Number of turns in the solenoid, 

n =50 turns/cm =5000 turns/m.  

If i is the current in the solenoid, then 

B =µₒni 

→i =B/µₒn

   =2.5x10⁻³/(4π*10⁻⁷*5000) A 

   =0.4 A.    


(b) Number of turns in the solenoid, 

n =50 turns/cm =5000 turns/m.

   After the core is inserted, the magnetic field inside the solenoid becomes, B =2.5 T. Now, 

B =µₒ(H+I) 

→H+I =B/µₒ 

→I =B/µₒ -H 

     =B/µₒ -ni

  =2.5/(4π*10⁻⁷) -5000*0·4 A/m

  =2.0x10⁶ -2000 A/m

  ≈2.0x10⁶ A/m.    


(c) Suppose there gets a pole strength m developed in the core. Area of the cross-section of the core A =4 cm² =4x10⁻⁴ m². The magnetization I of the core is given as, 

I = m/A  

→m = IA

   = 2.0x10⁶*4x10⁻⁴ A-m  

   = 800 A-m.        




 

    4. A bar magnet of length 1 cm and cross-sectional area 1.0 cm² produces a magnetic field of 1.5x10⁻⁴ T at a point in end-on position at a distance of 15 cm away from the center. (a) Find the magnetic moment M of the magnet. (b) Find the magnetization I of the magnet. (c) Find the magnetic field B at the center of the magnet.     


ANSWER: (a) Let the magnetic moment of the bar magnet = M. Distance of the point from the magnet in end-on position, r = 15.0 cm =0.15 m. The field is given as,

B =(µₒ/4π)*2M/r³

→M =½(4π/µₒ)Br³ 

 =½(10⁷)*1.5x10⁻⁴*(0.15)³ A-m² 

 =2.50 A-m².  


(b) Volume of the magnet,

  V =(1.0 cm²)*(1.0 cm) 

   =1.0 cm³

   =1.0x10⁻⁶ m³  

Magnetization of the magnet, 

I =M/V

  =2.50/(1.0x10⁻⁶) A/m 

  =2.5x10⁶ A/m.  


(c) Magnetic moment M =m*2l 

→m =M/2l =2.5/0.01 A-m 

→m =250 A-m.

Magnetic intensity H due to a magnetic pole of pole strength m,

H =m/(4πr²) 

The intensity at the center will be equal and in the same direction, hence total magnetic intensity at the center of the magnet, 

H =2m/(4πr²). 

Its direction will be from north to south pole of the magnet. The direction of magnetization I will be along the direction of the magnetic moment of the magnet which is from south pole to north pole of the magnet. So the directions of H and I are opposite to each other.

The net magnetic field at the center will be 

B =µₒ(H+I), 

{with appropriate directions}

→B  = -µₒ*2m/(4πr²) +µₒI 

  = -(µₒ/4π)*2m/r² +µₒI 

r =0.5 cm =0.005 m, hence  

B = -10⁻⁷*2*250/(0.005)²+4π*10⁻⁷*2.5x10⁶ T

  = -2+3.14 T   

  =1.14 T 

            

          

 



 

    5. The susceptibility of annealed iron at saturation is 5500. Find the permeability of annealed iron at saturation.    


ANSWER: Susceptibility of annealed iron at saturation, χ =5500.

Permeability at saturation is,

µ =µₒ(1+χ)

  =4πx10⁻⁷(1+5500)

  =6.9x10⁻³

 



 

    6. The magnetic field B and the magnetic intensity H in a material are found to be 1.6 T and 1000 A/m respectively. Calculate the relative permeability µᵣ and the susceptibility χ of the material.    


ANSWER: B =µH, where µ is the permeability of the material. Hence,  

µ =B/H

  =1.6/1000

  =0.0016 

Relative permeability µᵣ =µ/µₒ

→µᵣ =0.0016/(4πx10⁻⁷)

    =1.3x10³


Now the relation between susceptibility of the material and permeability is 

µ =µₒ(1+χ)

Where χ is the susceptibility.

0.0016 =4πx10⁻⁷(1+χ)

→1+χ =1.3x10³

→χ ≈ 1.3x10³.


             



 

    7. The susceptibility of magnesium at 300 K is 1.2x10⁻⁵. At what temperature will the susceptibility increase to 1.8x10⁻⁵?     


ANSWER: The susceptibility of a paramagnetic material is inversely proportional to the absolute temperature. Hence,  

χ/χ' =T'/T

→T' =χT/χ'

   =1.2x10⁻⁵*300/1.8x10⁻⁵

   =200 K. 




 

    8. Assume that each iron atom has a permanent magnetic moment equal to 2 Bohr magnetons (1 Bohr magneton equals 9.27x10⁻²⁴ A-m²). The density of atoms in iron is 8.52x10²⁸ atoms/m³. (a) Find the maximum magnetization I in a long cylinder of iron. (b) Find the maximum magnetic field B on the axis inside the cylinder.    


ANSWER: (a) Magnetization I =M/V 

Hence the maximum magnetization, taking one m³ of iron, 

I =2*9.27x10⁻²⁴*8.52x10²⁸ A/m 

  =1.58x10⁶ A/m. 


(b)  At this magnetization, the magnetic field, 

B =µₒ(H+I)

Since it is a long cylinder, end effects may be taken as zero. 

So, B =µₒI   

       =4πx10⁻⁷*1.58x10⁶ T 

      =1.98 T =2.0 T.    



 

    9. The coercive force for a certain permanent magnet is 4.0x10⁴ A/m. The magnet is placed inside a long solenoid of 40 turns/cm and a current is passed in the solenoid to demagnetize it completely. Find the current.  


ANSWER: Given, n =40 turns/cm

→n =4000 turns/m

Coercive force, 

H =4.0x10⁴ A/m

H =ni

→i =H/n

  = 4.0x10⁴/4000 A

  = 10 A.         

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Links to the Chapters





CHAPTER- 34- Magnetic Field

CHAPTER- 29- Electric Field and Potential











CHAPTER- 28- Heat Transfer

OBJECTIVE -I







EXERCISES - Q51 to Q55


CHAPTER- 27-Specific Heat Capacities of Gases

CHAPTER- 26-Laws of Thermodynamics


CHAPTER- 25-CALORIMETRY

Questions for Short Answer

OBJECTIVE-I

OBJECTIVE-II


EXERCISES - Q-11 to Q-18


CHAPTER- 24-Kinetic Theory of Gases







CHAPTER- 23 - Heat and Temperature






CHAPTER- 17 - Light Waves




CHAPTER- 14 - Fluid Mechanics



CHAPTER- 13 - Fluid Mechanics


CHAPTER- 12 - Simple Harmonic Motion








CHAPTER- 11 - Gravitation




CHAPTER- 10 - Rotational Mechanics




CHAPTER- 9 - Center of Mass, Linear Momentum, Collision


CHAPTER- 8 - Work and Energy

Click here for → Question for Short Answers

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → Exercises (1-10)

Click here for → Exercises (11-20)

Click here for → Exercises (21-30)

Click here for → Exercises (31-42)

Click here for → Exercise(43-54)

CHAPTER- 7 - Circular Motion

Click here for → Questions for Short Answer 

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → EXERCISES (11-20)

Click here for → EXERCISES (21-30)

CHAPTER- 6 - Friction

Click here for → Questions for Short Answer

Click here for → OBJECTIVE-I

Click here for → Friction - OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → Exercises (11-20)

Click here for → EXERCISES (21-31)

For more practice on problems on friction solve these- "New Questions on Friction".

---------------------------------------------------------------------------------

CHAPTER- 5 - Newton's Laws of Motion


Click here for → QUESTIONS FOR SHORT ANSWER

Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)

-------------------------------------------------------------------------------

CHAPTER- 4 - The Forces

The Forces-

"Questions for short Answers"    


Click here for "The Forces" - OBJECTIVE-I


Click here for "The Forces" - OBJECTIVE-II


Click here for "The Forces" - Exercises


--------------------------------------------------------------------------------------------------------------

CHAPTER- 3 - Kinematics - Rest and Motion

Click here for "Questions for short Answers"


Click here for "OBJECTIVE-I"


Click here for EXERCISES (Question number 1 to 10)


Click here for EXERCISES (Question number 11 to 20)


Click here for EXERCISES (Question number 21 to 30)


Click here for EXERCISES (Question number 31 to 40)


Click here for EXERCISES (Question number 41 to 52)


CHAPTER- 2 - "Physics and Mathematics"

Click here for "Questions for Short Answers"


Click here for "OBJECTIVE-II"

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