Thursday, December 23, 2021

H C Verma solutions, PERMANENT MAGNETS, Chapter-36, EXERCISES, Q1 to Q10, Concepts of Physics, Part-II

Permanent Magnets


EXERCISES, Q1 - Q10


    1.  A long bar magnet has a pole strength of 10 A-m. Find the magnetic field at a point on the axis of the magnet at a distance of 5 cm from the north pole of the magnet.   


ANSWER: Pole strength, m =10 A-m. Distance of the point from the pole, r =5 cm =0.05 m. Hence the magnetic field, 

B =(µₒ/4π)*m/r² 

 =(10⁻⁷)*10/(0.05)² T

 =4x10⁻⁴ T 

Outward along the axis from the north pole. 

     




 

    2.  Two long bar magnets are placed with their axes coinciding in such a way that the north pole of the first magnet is 2 cm from the south pole of the second. If both the magnets have a pole strength of 10 A-m, find the force exerted by one magnet on the other.   


ANSWER: Pole strength m =10 A-m, r =2 cm =0.02 m. The field at 2 cm from a pole, 

B = (µₒ/4π)(m/r²)

 =(10⁻⁷)*10/0.02² T 

 =2500x10⁻⁶ T 

 =2.5x10⁻³ T

Hence the force exerted on other magnet's pole, 

F =mB 

  =10*2.5x10⁻³ N

  =2.5x10⁻² N.         



 

    3.  A uniform magnetic field of 0.20x10⁻³ T exists in space. Find the change in the magnetic scalar potential as one moves through 50 cm along the field.  


ANSWER: The strength of the uniform magnetic field B =0.20x10⁻³ T. The change in the magnetic scalar potential is given as

V(r') -V(r) =-∫B.dr

  =-B∫dr

{Because the direction of Band dr is the same and B is uniform. The limit of integration is from r =0 to r' =50 cm =0.50 m} 

  =-Br'   

  =-0.20x10⁻³*0.50 T-m 

  =-0.1x10⁻³ T-m.  

So the magnetic scalar potential decreases by 0.1x10⁻³ T-m.    



 

    4.  Figure (36-E1) shows some of the equipotential surfaces of the magnetic scaler potential. Find the magnetic field B at a point in the region.   
The figure for Q-4


ANSWER: The direction of the magnetic field in the region at any point will be perpendicular to the equipotential surfaces. The strength of the field is, 

B =-dV/dl  

=-(-0.1x10⁻⁴ T-m)/(10xSin30°/100 m)

=1.0x10⁻⁴/(½) T 

=2.0x10⁻⁴ T    



 

    5.  The magnetic field at a point, 10 cm away from a magnetic dipole, is found to be 2.0x10⁻⁴ T. Find the magnetic moment of the dipole if the point is (a) in an end-on position of the dipole and (b) in a broadside-on position of the dipole.    


ANSWER: (a) In an end-on position of the dipole, the magnetic field is given as,   

B =(µₒ/4π)(2M/r³), where Mis the magnetic moment of the dipole. 

So,

2x10⁻⁴ =(10⁻⁷){2M/(0.10)³} 

→M =10⁷*10⁻⁴*10⁻³ A-m² =1.0 A-m²


(b) In the broadside-on position of a dipole, the magnetic field is given as,  

B =(µₒ/4π)(M/r³), so

2x10⁻⁴ =10⁻⁷*M/(0.10)³

→M =2x10⁻⁴*10⁷*10⁻³ A-m² 

   =2.0 A-m².          



 

    6.  Show that the magnetic field at a point due to a magnetic dipole is perpendicular to the magnetic axis if the line joining the point with the center of the dipole makes an angle of tan⁻¹(√2) with the magnetic axis.  


ANSWER: See the figure below. θ is the angle between the line joining the point with the center of the dipole and the magnetic axis while α is the angle between magnetic field B and the line joining the point with the center of the dipole.   
Figure for Q-6

When the magnetic field B is perpendicular to the magnetic axis the sum, 

α+θ =π/2 

→α =π/2 -θ  

Also in general, tan α =(tanθ)/2

→tan(π/2 -θ) =(tanθ)/2

→Cotθ =(tanθ)/2

→tan²θ =2

→tanθ =√2 

→θ =tan⁻¹(√2).      




    7.  A bar magnet has a length of 8 cm. The magnetic field at a point at a distance of 3 cm from the center in the broadside-on position is found to be 4x10⁻⁶ T. Find the pole strength of the magnet.  


ANSWER: Length of the bar magnet, 

2l =8 cm =0.08 m

→l =0.04 m.   

Distance of the point from center of the magnet, d = 3 cm = 0.03 m. 

The magnetic field B =4x10⁻⁶ T, (given).

But B = (µₒ/4π)M/(d²+l²)³'² 

→M =B(d²+l²)³'²/(µₒ/4π) 

  =4x10⁻⁶(0.03²+0.04²)³'²*10⁷ A-m²

  =40*0.05³ A-m² 

  =0.005 A-m² 

→m*2l =0.005 A-m² 

→m =0.005/0.08 A-m

      =0.0625 A-m       

      =6.25x10⁻² A-m.



 

    8.  A magnetic dipole of magnetic moment 1.44 A-m² is placed horizontally with the north pole pointing towards the north. Find the position of the neutral point if the horizontal component of the earth's magnetic field is 18 µT.   


ANSWER: Horizontal component of earth's magnetic field, BH =18 µT.  

The neutral point will be at a point distance d away in the broad-on position where the magnetic field due to the dipole (B') and BH are equal in magnitude but opposite in direction. See the diagram below.
Diagram for Q-8

    Magnetic field in broad-on position of a dipole at distance d from the center of the magnet,

   B' =(µₒ/4π)*M/d³ 

→d³ =(µₒ/4π)*M/B' 

   =(10⁻⁷)*1.44/(18x10⁻⁶) 

  =0.008 m³ 

→d =0.20 m =20 cm     

Such points will be all around the dipole making a circle. So the neutral point is at a distance of 20 cm in the plane bisecting the dipole. 





 

    9.  A magnetic dipole of magnetic moment 0.72 A-m² is placed horizontally with the north pole pointing towards the south. Find the position of the neutral point if the horizontal component of the earth's magnetic field is 18 µT.   


ANSWER: The direction of horizonal component of the earth's magnetic field is towards north. The direction of  magnetic field due to the dipole opposite in direction i.e. towards the south will be in end-on position, south of the dipole. The neutral point will be at that point south of the dipole where the magnitude is equal to horizontal componenet of the earth's magnetic field but opposite in direction. 

B' =(µₒ/4π)*(2M/d³)

→d³ =(µₒ/4π)*(2M/B')

  =(10⁻⁷)*2*0.72/(18x10⁻⁶) m³ 

 =0.008 m³

→d =0.20 m =20 cm.       

So the neutral point in this case will be 20 cm south of the dipole. 



 


    10.  A magnetic dipole of magnetic moment 0.72√2 A-m² is placed horizontally with the north pole pointing towards the east. Find the position of the neutral point if the horizontal component of the earth's magnetic field is 18 µT.   


ANSWER: Since the dipole is in east west position, the magnetic field opposite to the horizontal component of earth's magnetic field will be at an angle θ from the axis of the dipole. Suppose the magnitude of the magnetic field due to the dipole that is equal to BH at the neutral point at distance d from it. See the diagram below. 
Diagram for Q-10

   In such case α+θ =90°, 

→α = 90°-θ. Also,

tanα =(tanθ)/2

→tan(90°-θ) =(tanθ)/2

→cotθ =(tanθ)/2

→1/tanθ =(tanθ)/2

→tan²θ =2

→tanθ =√2.

So, θ =tan⁻¹√2.

The magnitude of the magnetic field,

B' =(µₒ/4π)M√(1+3.cos²θ)/d³ ----(i) 


cosθ = 1/secθ =1/√(1+tan²θ)

  =1/√(1+2) =1/√3.

And B' =BH =18 µT ==18x10⁻⁶ T.

Now from (i) 

d³=(10⁻⁷)*0.72√2*√(1+1)/18x10⁻⁶ m³

  =0.008 m³.

→d =0.20 m =20 cm.

So the neutral point is at 20 cm from the center of the dipole and it is tan⁻¹√2 south of east.

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Links to the Chapters



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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