Sunday, October 31, 2021

H C Verma solutions, MAGNETIC FIELD DUE TO A CURRENT, Chapter-35, EXERCISES, Q21_Q30, Concepts of Physics, Part-II

Magnetic Field Due to a Current


EXERCISES, Q21 to Q30


    21.  A wire of length l is bent in the form of an equilateral triangle and carries an electric current i. (a) Find the magnetic field B at the center. (b) If the wire is bent in the form of a square, what would be the value of B at the center?      


ANSWER: (a) The direction of the magnetic field at the center of the equilateral triangle will be perpendicular to the plane of the triangle due to each of the sides. Hence the magnitude of the net magnetic field at the center will be

B =3*µₒia/{2πd√(a²+4d²)} 
Diagram for Q-21

    Here, a =l/3,

  d= OD = ⅓AD =⅓(l/3)*sin60 =√3l/18

So,

B=3µₒil/{3*2π√3l√(l²/9 +12l²/18²)/18}

 =µₒi/{2√3πl√48/18²} 

 =µₒi*18²/2*4(√3)²πl 

 =µₒi*18²/24πl 

 =13.5µₒi/πl  

 =27µₒi/2πl.  


   (b) The direction of the magnetic field at the center of the square wire will still be the same i.e. perpendicular to the plane of the square. The magnitude of the magnetic field will be 

B =4*µₒia/{2πd√(a²+4d²)} 

here, a =l/4, d =½*l/4 =l/8, so,

B =(4*µₒil/4)/{2π(l/8)√(l²/16+4l²/64)}

 =µₒi/{πl√2/16} 

 =16µₒi/√2πl 

 =8√2µₒi/πl.               





 

 

    22.  A long wire carrying a current i is bent to form a plane angle α. Find the magnetic field B at a point on the bisector of this angle situated at a distance x from the vertex.    


ANSWER: The distance of given point P from each side of the angle, 

d =xSin(α/2) 
Diagram for Q-22

 The direction and magnitude of the magnetic field at P will be the same due to each of the wires. The direction will be perpendicular to the plane of the angle. Hence the magnitude of the magnetic field at P will be double that due to one side wire. 

B =2*µₒi(cosθ₁ -cosθ₂)/4πd

Here θ₁ =α/2, θ₂ =π, so,

B=2µₒi.{cos(α/2)+1}/4πxSin(α/2)

=µₒi{2cos²(α/4)}/2πx{2Sin(α/4)cos(α/4)}

=µₒi{cos(α/4)/sin(α/4)}/2πx

=µₒiCot(α/4)/2πx. 





 


    23.  Find the magnetic field B at the center of a rectangular loop of length l and width b, carrying a current i.    


ANSWER: The direction of the magnetic field due to each side of the rectangle will be the same at the point i.e. perpendicular to the plane of the rectangle. 
Diagram for Q-23

    The magnitude of the magnetic field due to each width will be the same and also due to each length will also be the same. 

    Hence the net magnetic field at the center, 

B =2µₒil/{2πd√(l²+4d²)}+2µₒib/{2πd'√(b²+4d'²)}

=µₒil/{π½b√(l²+4b²/4)}+µₒib/{π½l√(b²+4l²/4)}

={2µₒi/π√(l²+b²)}(l/b +b/l) 

=2µₒi(l²+b²)/{πlb√(l²+b²)} 

=2µₒi√(l²+b²)/πlb.          




 

 

 

    24.  A regular polygon of n sides is formed by bending a wire of total length 2πr which carries a current i. (a) Find the magnetic field B at the center of the polygon. (b) By letting n→∞, deduce the expression for the magnetic field at the center of a circular current.     


ANSWER: (a) Length of each side of the polygon a =2πr/n. The angle is subtended at the center by each side, ß =2π/n. Half of this angle, ß/2 =π/n. If the distance of the center from each side = d. 

    Then tan(ß/2) =(a/2)/d =a/2d 

→d =a/2tan(ß/2) =a/2tan(π/n) 

The direction of the magnetic field at the center due to each side is the same i.e. perpendicular to the plane of the polygon. Hence the net magnitude of the magnetic field at the center, 

B=n*µₒia/{2πd√(a²+4d²)} 

=nµₒi(2πr/n)/{2π(a/2tan(π/n))√(a²+a²/tan²(π/n)} 

=µₒir{2tan²(π/n)/{a²√(1+tan²(π/n)}

=2µₒir.tan²(π/n)/{a²sec(π/n)} 

=2µₒirSin(π/n)tan(π/n)/a² 

=2µₒirSin(π/n)tan(π/n)/(2πr/n)² 

=µₒin²Sin(π/n).tan(π/n)/2π²r. 


(b) For a circle, n→∞. So π/n →0. The expression can be written as,  

B =µₒi{sin(π/n)/(π/n)}²/2rcos(π/n) 

when π/n →0, cos(π/n) ≈1 and also the expression 

sin(π/n)/(π/n) → 1. Thus,

B =µₒi/2r.              







 

    25.  Each of the batteries shown in figure (35-E8) has an emf equal to 5 V. Show that the magnetic field B at the point P is zero for any set of values of the resistances.     
Figure for Q-25


ANSWER: Let us assume the currents and resistances as shown in the diagram below. 
Diagram for Q-25

    Applying Kirchoff's loop law in the left loop anticlockwise. 

i₁r₁+5 +i₂r₂-5 = 0

→i₁r₁ +i₂r₂ =0

Similarly, i₂r₂ +i₃r₃ =0

But i₃+i₁ =i₂

So the first one is now,

i₁r₁ +r₂i₃+r₂i₁ =0

→(r₁+r₂)i₁ +r₂i₃ =0 -------- (i)

The second one becomes,

r₂i₃+r₂i₁+i₃r₃ =0

→r₂i₁+(r₂+r₃)i₃ =0

→{r₂²/(r₂+r₃)}i₁ +r₂i₃ =0 ------(ii)

Subtracting (ii) from (i), we get

{(r₁+r₂) -r₂²/(r₂+r₃)}i₁ =0

→i₁ =0

Putting it in (i), we get. i₃ =0.

Now i₂ =i₃+i₁ =0.

So whatever be the resistances in the circuit, there will be no current. Hence the magnetic field at any point (and hence also point P) will be zero.


   





    26.  A straight long wire carries a current of 20 A. Another wire carrying equal current is placed parallel to it. If the force acting on a length of 10 cm of the second wire is 2.0x10⁻⁵ N, what is the separation between them?    


ANSWER: Given, i₁ =i₂ =20 A. The force per unit length of the wire is given as =µₒi₁i₂/2πd, where d is the separation of the wire. Hence the force on 10 cm of the second wire,

0.10*µₒ*(20)²/2πd =2x10⁻⁵

→d =(4πx10⁻⁷/2π)*400x10⁴/2 m

 =40 m.     






 

    27.  Three coplanar parallel wires, each carrying a current of 10 A along the same direction, are placed with a separation of 5.0 cm between the consecutive ones. Find the magnitude of the magnetic force per unit length acting on the wires.     


ANSWER: Parallel wires carrying currents in the same direction attract each other. Since the separations and the currents are the same between the middle and side wires, each side wire will attract the middle one towards them with equal force. So the forces per unit length acting on the middle wire is equal and opposite. Thus net force is zero. 

   The force per unit length on a side wire is due to another side wire and the middle wire. Their directions are also the same i.e. towards the middle wire. Hence the net magnetic force per unit length on a side wire 

=µₒi²/2πd +µₒi²/{2π(2d)} 

=1.5µₒi²/2πd 

=1.5*(2x10⁻⁷)*10²/0.05 N 

=6.0x10⁻⁴ N.        





  


    28.  Two parallel wires separated by a distance of 10 cm carry currents of 10 A and 40 A along the same direction. Where should a third current be placed so that it experiences no magnetic force?   


ANSWER: Since the parallel wires carry current in the same direction the third current should be put in between them so that the force is equal and opposite. Let the third current i is placed at x cm distance from the 10 A wire and 10-x cm from the 40 A wire. Force per cm length on it due to 10 A wire, 

=µₒ*10i/2πx N 

Due to 40 A wire, 

=µₒ*40i/2π(10-x) N 

Equating these two we get, 

1/x =4/(10-x) 

→10-x =4x

→5x =10

→x =2 cm.  

So the third current should be placed 2 cm from the 10 A current and 10-2 =8 cm from the 40 A current.             






 

    29.  Figure (35-E9) shows a part of an electric circuit. The wires AB, CD, and EF are long and have identical resistances. The separation between neighboring wires is 1.0 cm. The wire AE and BF have negligible resistances and the ammeter reads 30 A. Calculate the magnetic force per unit length of AB and CD.      
Figure for Q-29


ANSWER: Since three parallel wires have equal resistances, the total current of 30 A will be equally divided and each wire will carry 10 A. Wires AE and BF are perpendicular hence will not affect the three wires. 

  Magnetic force per unit length of AB due to middle wire, 

=µₒi²/2πd and due to bottom wire, 

=µₒi²/4πd. Both forces downwards. Hence net force per unit length of AB

=µₒi²/2πd +µₒi²/4πd  

=3µₒi²/4πd 

=3*(1x10⁻⁷)*10²/(0.01) N/m 

=3.0x10⁻³ N/m. Downward.  


   The force per unit length on the middle wire CD due to AB and EF will be equal and opposite, hence net foce zero.           




  

  

    30.  A long straight wire is fixed horizontally and carries a current of 50.0 A. A second wire having linear mass density 1.0x10⁻⁴ kg/m is placed parallel to and directly above this wire at a separation of 5.0 mm. What current should this second wire carry such that the magnetic repulsion can balance its weight?    


ANSWER: To have a magnetic repulsion the current in the second wire should be in opposite direction. Let the magnitude of the current be i. d =5 mm =0.005 m. The magnetic force of repulsion per m =1x10⁻⁴*9.8 N =9.8x10⁻⁴ N.  

So, µₒ*50i/(2π*0.005) =9.8x10⁻⁴ 

→2x10⁻⁷*i =9.8x10⁻⁸ 

→i =4.9x10⁻¹ A 

→i =0.49 A.        

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

Books for Different Entrance Exams


--------------------------------------------------- Buy Home Furnishing

Click here for all links → kktutor.blogspot.com 

===<<<O>>>===


My Channel on YouTube  →  SimplePhysics with KK


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"

Posted by at No comments:
Labels: , , , ,
Subscribe to: Posts (Atom)