Monday, December 27, 2021

H C Verma solutions, PERMANENT MAGNETS, Chapter-36, EXERCISES, Q21 to Q25, Concepts of Physics, Part-II

Permanent Magnets


EXERCISES, Q21 - Q25


    21.  A bar magnet takes π/10 seconds to complete one oscillation in an oscillation magnetometer. The moment of inertia of the magnet about the axis of rotation is 1.2x10⁻⁴ kg-m² and the earth's horizontal magnetic field is 30 µT. Find the magnetic moment of the magnet.   


ANSWER: Time period, T = π/10 s. The moment of inertia of the magnet, I =1.2x10⁻⁴ kg-m². BH =30µT. If the magnetic moment of the magnet be M, then,  

T =2π√(I/MBH) 

→M =4π²I/BHT² 

 =4π²*1.2x10⁻⁴/(30x10⁻⁶*π²/100) 

 =1.6x10³ A-m²

 =1600 A-m².       





 

    22.  The combination of two bar magnets makes 10 oscillations per second in an oscillation magnetometer when like poles are tied together and 2 oscillations per second when unlike poles are tied together. Find the ratio of the magnetic moments of the magnets. Neglect any induced magnetism.   


ANSWER: Let the magnetic moments of the magnets be M and M'. When like poles are tied together the net magnetic moment =M +M'. When unlike poles are tied together, the net magnetic moment =M -M'. 

   In the first case time period, T =1/10 s =0.10 s. In the second case time period, T' =1/2 s =0.50s. 

The relation between magnetic moment, time period, the moment of inertia of the magnet, and BH is given as 

T =2π√(I/MBH). 

→M =4π²I/BHT²

 Hence in the first case,

M+M' =4π²I/BHT² ---------- (i) 

   Where I is the combined moment of inertia of the magnets.  

In the second case,

M-M' =4π²I/BHT'² ----------- (ii) 

Dividing (i) by (ii), we get 

(M+M')/(M-M') =T'²/T² =(0.5/0.1)² =25/1

     From componendo and dividendo rule

(M+M'+M-M')/(M+M'-M+M')=(25+1)/(25-1)

→2M/2M' =26/24 

→M/M' =13:12.                




 

    23.  A short magnet oscillates in an oscillation magnetometer with a time period of 0.10 s where the earth's horizontal magnetic field is  24 µT. A downward current of 18 A is established in a vertical wire placed 20 cm east of the magnet. Find the new time period.   


ANSWER: Since the current in the wire is downwards, the circular magnetic field around it will be clockwise when viewed from above. Also, since the magnet is short, the magnetic field, B' due to the current will be taken towards the north. 

 B' =µₒi/(2πd) 

 =(µₒ/2π)*18/0.2 

 =180µₒ/4π T 

 =180*10⁻⁷ T 

 =18 µT.     
Diagram for Q-23

Now the total magnetic field in the north direction B" =B'+BH  

 =18 +24 T =42 µT. 

The time period of the oscillation without the current, 

T =2π√(I/MBH)   and the time period with the current, 

T' =2π√(I/MB").  

Now, T'/T =√(BH/B") 

→T' =T√(24/42)

  =0.10*0.76 s =0.076 s. 

So the new time period is 0.076 s.           




 

    24.  A bar magnet makes 40 oscillations per minute in an oscillation magnetometer. An identical magnet is demagnetized completely and is placed over the magnet in the magnetometer. Find the time taken for 40 oscillations by the combination. Neglect any induced magnetism.     


ANSWER: The time period in the first case, T =1/40 minutes =1.5 s. If the moment of inertia of the magnet is I, then in the second case the combined moment of inertia =2I. Let the magnetic moment of the magnet and the horizontal component of the earth's magnetic field be M and BH respectively.  

 T =2π√(I/MBH)  ----- --- (i)

If the new time period is T' then, 

T' =2π√(2I/MBH) ----- --- (ii)  

Dividing (ii) by (i), 

T'/T =√2 

→T' =√2T =1.5√2 s 

It is the time for one oscillation, hence the time taken to perform 40 oscillations =40*1.5√2 s =60√2 s =√2 minutes.         




  

    25.  A short magnet makes 40 oscillations per minute when used in an oscillation magnetometer at a place where the earth's horizontal magnetic field is 25 µT. Another short magnet of magnetic moment 1.6 A-m² is placed 20 cm east of the oscillating magnet, Find the new frequency of oscillation if the magnet has its north pole (a) towards the north and (b) towards the south.   


ANSWER: The oscillating magnet is at the broadside-on position of the short magnet placed 20 cm east of it. The magnetic field due to the second magnet at the position of the first magnet, 

B' =(µₒ/4π)M/d³

 =10⁻⁷*1.6/0.2³ T 

 =20x10⁻⁶ T 

 =20 µT.  

The time period of oscillation before placing the magnet, T =60/40 s =1.5 s. 

So, T =2π√(I/MBH) 

 

(a) When the second magnet has its north pole towards the north, the direction of B' is towards the south i.e. opposite to BH. The net horizontal field at the first magnet, B" =BH-B' =25-20 µT

→B" =5 µT. 

So the time period now, 

T' =2π√(I/MB") 

Hence, 

T'/T =√(BH/B")  ---- ----- (i)

→T' =1.5√(25/5) =3.35 s 

Oscillation per minute =60/3.35 ≈18.     

(b) When the second magnet has its north pole towards the south the direction of B' at the position of the first magnet is towards the north i.e. same as BH. Hence the net horizontal magnetic field at the oscillating magnet, 

B" =BH+B' =25+20 =45 µT. 

The new time period T' is given from (i) as, 

T' =1.5√(25/45) =1.12 s 

Hence oscillation per minute =60/1.12 =54.            


 

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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

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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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