Saturday, December 25, 2021

H C Verma solutions, PERMANENT MAGNETS, Chapter-36, EXERCISES, Q11 to Q20, Concepts of Physics, Part-II

Permanent Magnets


EXERCISES, Q11 - Q20


    11.  The magnetic moment of the assumed dipole at the earth's center is 8.0x10²² A-m². Calculate the magnetic field B at the geomagnetic poles of the earth. The radius of the earth is 6400 km.    


ANSWER: The geomagnetic poles are at 6400 km from the center of the earth on the magnetic axis. So we have to find the magnetic field on the end-on position of the dipole at a distance r =6.4x10⁶ m. 

Hence the required field, 

B =(µₒ/4π)*2M/r³

 =2*(10⁻⁷)*8x10²²/(6.4x10⁶)³ T 

 =6.0x10⁻⁵ T

 =60x10⁻⁶ T

 =60 µT.  

     




 

    12.  If the earth's magnetic field has a magnitude 3.4x10⁻⁵ T at the magnetic equator of the earth, what would be its value at the earth's geomagnetic poles?   


ANSWER: The magnetic equator and the geomagnetic poles are at the same distance from the center of the earth that is equal to the distance of the earth's radius R. The magnetic equator is in a broadside-on position and the field is given as B =(µₒ/4π)M/R³. 

The poles are at the end-on position and the field is given as B' =(µₒ/4π)*2M/R³. 

 So the field at the poles is double the field at the equator. Given the magnitude of the field at the magnetic equator B=3.4x10⁻⁵ T. Hence the magnitude of the field at the geomagnetic poles,

 B' =2B 

  =2x3.4x10⁻⁵ T

  =6.8x10⁻⁵ T.       



 

    13.  The magnetic field due to the earth has a horizontal component of 26 µT at a place where the dip is 60°. Find the vertical component and the magnitude of the field.   


ANSWER: The horizontal component of the earth's magnetic field, 

BH =B.cosδ

Where B is the magnetic field of the earth and δ is the dip at that place. Hence the magnitude of the earth's field 

B =BH/cosδ

 =26/cos60° µT

 =26*2 µT

 =52 µT.

 

Also the vertical component of the magnetic field

BV =B.sinδ

 =52*sin60° µT

 =52*√3/2 µT

 =45 µT.     




 

    14.  A magnetic needle is free to rotate in a vertical plane which makes an angle of  60° with the magnetic meridian. If the needle stays in a direction making an angle of tan⁻¹(2/√3) with the horizontal, what would be the dip at that place?    


ANSWER: Angle between magnetic meridian and the plane, 

α =60°.

Apparent dip, δ' =tan⁻¹(2/√3) 

→tanδ' =(2/√3)

The true dip at that place δ is given as, 

cotδ' =cotδ*cosα

→Cotδ =cotδ'/cosα 

   =1/(tanδ'*cosα)       

   =1/(2/√3*cos60°) 

   =√3 

→tanδ =1/√3 = tan30°

→δ =30°.   


Otherwise, you can understand it as below:-

Effective horizontal component of the magnetic field

BH' = BHcos60° =BH/2  

Apparent dip δ' is given by 

tanδ' =Bv/BH' =2Bv/BH =2tanδ

(where δ is the true dip). 

→2/√3 =2tanδ

→tanδ =1/√3
→δ =30°.

    


 

    15.  The needle of a dip circle shows an apparent dip of 45° in a particular position and 53° when the circle is rotated through 90°. Find the true dip.   


ANSWER: Here, apparent dip δ' =45°. Suppose the plane of the dip circle makes an angle α with the magnetic meridian. Now if the true dip is δ then, 

cotδ' =cotδ cosα   --------(i)

When the dip circle is rotated through 90°, the angle between the plane of the dip circle and the meridian=90°-α. So the apparent dip now δ" is given as, 

 cotδ" =cotδ cos(90°-α) =cotδ sinα --(ii)

Squaring and adding (i) and (ii), 
cot²δ =cot²δ' +cot²δ" 
→cot²δ =cot²45°+cot²53°
         =1² +(3/4)² 
         =5²/4² 
→cotδ =5/4 
→tanδ =4/5 =tan39°
→δ = 39°.    

    



 

   16.  A tangent galvanometer shows a deflection of 45° when 10 mA of current is passed through it. If the horizontal component of the earth's magnetic field is BH = 3.6x10⁻⁵ T and the radius of the coil is 10 cm, find the number of turns in the coil.   


ANSWER: Radius of the coil r =0.10 m. Deflection θ =45°. Current i =10 mA =0.01 A. Horizontal component of earth's magnetic field BH =3.6x10⁻⁵ T. If n is the number of turns in the coil, then the current is given as,

 i =2rBH*tanθ/(µₒn)

→n =2*0.10*3.6x10⁻⁵/(4πx10⁻⁷*0.01)

 =0.573x10³

 =573. 



 

    17.  A moving-coil galvanometer has a 50-turn coil of size 2 cm x 2 cm. It is suspended between the magnetic poles producing a magnetic field of 0.5 T. Find the torque on the coil due to the magnetic field when a current of 20 mA passes through it.   


ANSWER: Number of turns, n =50. 

Area of the coil A =0.02x0.02 m² 

→A =4x10⁻⁴ m² 

Magnetic field between the magnetic poles,

B = 0.5 T 

Current in the coil, i =20 mA =0.02 A

Hence the torque on the coil, 

𝛤 =niAB 

  =50*0.02*4x10⁻⁴*0.5 N-m 

  =2x10⁻⁴ N-m.         



 

    18.  A short magnet produces a deflection of 37° in a deflection magnetometer in Tan-A position when placed at a separation of 10 cm from the needle. Find the ratio of the magnetic moment of the magnet to the earth's horizontal magnetic field.   


ANSWER: Deflection θ =37°. Distance of the magnet from the needle d =10 cm =0.10 m

In the Tan-A position, the needle is in the end-on position of the magnet. Hence the magnetic field produced by the magnet at the needle,

B =(µₒ/4π)*2Md/(d²-l²)²

Since the magnet is short, so l<<d, we can neglect l. Thus

B =(µₒ/4π)*2M/d³

In Tan-A position, B and BH are perpendicular to each other and,

B = BH tanθ

→(µₒ/4π)*2M/d³ =BH tanθ

→M/BH =½(4π/µₒ)d³tanθ

   =½(10⁷)*(0.10)³*tan37° A-m²/T

   =(½*¾)x10⁴ A-m²/T 

   =3.75x10³ A-m²/T.

 



 

    19.  The magnetometer of the previous problem is used with the same magnet in the Tan-B position. Where should the magnet be placed to produce a 37° deflection of the needle?   


ANSWER: In the Tan-B position, the needle is in a broadside-on position. The magnetic field by the magnet at needle is given as 

B =(µₒ/4π)M/(d²+l²)³'² 

For the short magnet we ignore l, thus

B =(µₒ/4π)M/d³

Since B and BH are perpendicular to each other,

B =BH tanθ

(µₒ/4π)M/d³ =BH tanθ

d³ =(µₒ/4π)(M/BH)/tan37°

 =(10⁻⁷)(3.75x10³)/(¾) m³

 =5x10⁻⁴ m³

 

→d =0.079 m =7.9 cm 

So the magnet should be placed at 7.9 cm from the center in Tan-B position to produce a deflection of 37° in the needle.




 

    20.  A deflection magnetometer is placed with its arm in the north-south direction. How and where should a short magnet having M/BH =40 A-m²/T be placed so that the needle can stay in any position?    


ANSWER: The horizontal component of the earth's magnetic field has a direction along the arm of the magnetometer i.e. from south to north. The condition to stay the needle in any position can be achieved if the net field in the horizontal direction is zero at the center and it can be done by putting the magnet in the north-south direction on the arm so that the north pole of the magnet faces the magnetic south direction. The needle is in the end-on position of the short magnet. The distance d is adjusted to have, 

B =BH 

but opposite in direction.

(µₒ/4π)*2M/d³ =BH

→d³ =2(µₒ/4π)(M/BH)

  =2(10⁻⁷)(40) m³

  =8x10⁻⁶ m³ 

d =2x10⁻² m = 0.02 m

→d =2 cm.     

So the short magnet should be placed 2.0 cm from the needle, north pole pointing towards the south. 

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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

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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"

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