Monday, November 15, 2021

H C Verma solutions, MAGNETIC FIELD DUE TO A CURRENT, Chapter-35, EXERCISES, Q41_Q50, Concepts of Physics, Part-II

Magnetic Field Due to a Current


EXERCISES, Q41 to Q50


    41.  Find the magnetic field B due to a semicircular wire of radius 10.0 cm carrying a current of 5.0 A at its center of curvature.      


ANSWER: If the magnetic field due to a very small length dl at the center is dB, then  

dB =(µₒi/4π)dlxr/r³  

Since dl and 'r' are at right angles, the magnitude  

dB =(µₒi/4π)dl/r² 

Now the total magnetic field at the center 

B =∫dB

 =(µₒi/4πr²)∫dl

 =(µₒi/4πr²)*πr

 =µₒi/4r. 

 =4πx10⁻⁷*5/(4*0.10) T

 =15.7x10⁻⁶ T

 =1.6x10⁻⁵ T.

        





 

    42.  A piece of wire carrying a current of 6.00 A is bent in the form of a circular arc of radius 10.0 cm, and it subtends an angle of 120° at the center. Find the magnetic field B due to this piece of wire at the center.      


ANSWER: In this case the ∫dl 

=(120°/360°)*2πr

=⅔πr

Hence B=(µₒi/4πr²)∫dl

→B =(µₒi/4πr²)*⅔πr

  =µₒi/6r

  =4πx10⁻⁷*6/(6*0.10) T

 =4πx10⁻⁶ T

 =12.6x10⁻⁶ T

 =1.26x10⁻⁵ T.

 




 

    43.  A circular loop of radius r carries a current i. How should a long, straight wire carrying a current 4i be placed in the plane of the circle so that the magnetic field at the center becomes zero?       


ANSWER: The direction of the magnetic field at the center of the loop will be coming towards the viewer if the current is anticlockwise. The direction of the magnetic field around a straight wire is anticlockwise if the current is coming towards the viewer. To keep the direction of the magnetic field due to the wire in the plane of the loop opposite at the center of the loop, the direction of the current in the wire and the nearest portion of the loop should be opposite. 
Diagram for Q-43


To make the net magnetic field zero at the center, both magnitudes should be equal. Hence, 

µₒi/2r = µₒ(4i)/2πd, where d is the distance of the long wire from the center of the loop. 

→d =4r/π.     




 

    44.  A circular coil of 200 turns has a radius of 10 cm and carries a current of 2.0 A. (a) Find the magnitude of the magnetic field B at the center of the coil. (b) At what distance from the center along the axis of the coil will the field B drop to half its value at the center? (∛4 =1·5874...).    


ANSWER: (a) n =200, radius a =10 cm =0.10 m, i =2.0 A. The magnitude of the magnetic field at the center of the loop 

 B =µₒni/2a 

 =(4πx10⁻⁷*200*2)/(2*0.10) T 

 =2.51x10⁻³ T  

 =2.51 mT 


(b) The magnetic field at a point distance d from the center along the axis of the loop is, 

B' =µₒnia²/{2(a²+d²)3/2} 

In this case, B' =B/2  

→µₒni/4a =µₒnia²/{2(a²+d²)3/2} 

→(a²+d²)3/2 = 2a³ 

→(a²+d²)³ =4a⁶ 

→a²+d² =a²∛4 

→d² =(1.5874 -1)a²

→d² =0.5874a² 

→d =0.766a =0.766*10 cm  

      =7.66 cm.                




 

    45.  A circular loop of radius 4.0 cm is placed in a horizontal plane and carries an electric current of 5.0 A in the clockwise direction as seen from above. Find the magnetic field (a) at a point 3.0 cm above the center of the loop (b) at a point 3.0 cm below the center of the loop.     


ANSWER: Since the current in the horizontal loop is in the clockwise direction when seen from above, the direction of the magnetic field in each case will be downward. Also, the points above and below the center of the loop are equidistant from the center, the magnitude of the magnetic field will also be the same. This magnitude, 

B =µₒia²/{2(a²+d²)¹∙⁵} 

 =(4πx10⁻⁷*5.0*0.04²)/{2(0.04²+0.03²)¹∙⁵} T 

 =3.2πx10⁻⁹/{2(0.05²)¹∙⁵} T

 =1.6πx10⁻⁹/0.05³ T 

 =4.0x10⁻⁵ T. Downwards in both the cases.    




 


 

    46.  A charge of 3.14x10⁻⁶ C is distributed uniformly over a circular ring of radius 20.0 cm. The ring rotates about its axis with an angular velocity of 60 rad/s. Find the ratio of the electric field to the magnetic field at a point on the axis at a distance of 5.00 cm from the center.     


ANSWER: In one revolution of the ring, 2π radians are covered. So per second, 60/2π revolutions of the ring will take place. Hence the equivalent current in the ring 

i =Charge passing through a section per second 

=60*3.14x10⁻⁶/2π A

=3x10⁻⁵ A  

The radius of the ring, a =0.20 m  

The electric field at this point, 

E =(1/4πεₒ)*Qd/(a²+d²)¹·⁵  

The magnetic field at this point 

B =µₒia²/{2(a²+d²)¹·⁵}

 =(1/4πεₒc²)4πia²/{2(a²+d²)¹·⁵} 

{Substituting for µₒ} 

Hence, 

E/B=2Qdc²/4πia² 

   =Qdc²/2πia² 

=3.14x10⁻⁶*0.05*(3x10⁸)²/(2π*3x10⁻⁵*0.20²) m/s

=1.88x10¹⁵ m/s.   

  The unit of this ratio,

E/B = (N/C)/(N/A-m) 

  = A-m/C 

  = (C/s)*m/C

  =m/s




 

    47.  A thin, long, hollow cylindrical tube of radius r carries current 'i' along its length. Find the magnitude of the magnetic field at a distance r/2 from the surface (a) inside the tube (b) outside the tube.     


ANSWER: (a) Consider an amperes circular loop of radius r/2 with its center on the axis of the tube. 
Diagram for Q-47

The circulation, 

∮B.dl =µₒi

→B*2πr/2 = 0, since current inside this loop is zero.

→B =0.

So the magnitude of the magnetic field at r/2 distance inside the tube is zero.


(b) Taking the ampere circular loop through the point outside the tube at r/2 distance from the surface and the center on the axis of the tube, the circulation,

∮B.dl =µₒi

→B*2π(3r/2) =µₒi

→B =µₒi/(3πr).      




 

    48.  A long cylindrical tube of inner and outer radii a and b carries a current 'i' distributed uniformly over its cross-section. Find the magnitude of the magnetic field at a point (a) just inside the tube (b) just outside the tube.      


ANSWER: (a) Take an ampere circular loop with its center on the axis of the cylindrical tube. The circulation ∮B.dl  along this loop is,

∮B.dl =µₒi

→B*2πa =µₒi

{Here enclosed current, i = 0}

→2πBa =0

→B =0.


(b) Similarly taking an ampere circular loop just outside the tube with its center on the axis of the tube. The circulation ∮B.dl along it is,

∮B.dl =µₒi

→B*2πb =µₒi, {Enclosed current =i}

→B =µₒi/2πb.

 




 

    49.  A long cylindrical wire of radius b carries a current 'i' distributed uniformly over its cross-section. Find the magnitude of the magnetic field at a point inside the wire at a distance 'a' from the axis.     


ANSWER: When we take an ampere circular loop through the given point with its center on the axis, the portion of the current through this loop,   

i' =i*(πa²/(πb²) =ia²/b²  

The circulation ∮B.dl over this loop is,

∮B.dl = µₒi'

→B*2πa =µₒia²/b²

→B =µₒia/2πb².      




 

    50.  A solid wire of radius 10 cm carries a current of 5.0 A distributed uniformly over its cross-section. Find the magnetic field B at a point at a distance (a) 2 cm  (b) 10 cm and (c) 20 cm away from the axis. Sketch a graph of B versus x for 0<x<20 cm.     


ANSWER: Taking the ampere circular loop through a point with the center of this loop on the axis of the wire. Due to symmetry, the magnitude of B at each point of this loop will be the same. The current enclosed by this loop when the point is inside the wire,

i' =i*(πx²)/(πr²)

 =ix²/r²

 The circulation ∮B.dl over this loop

∮B.dl =µₒi'

→B*2πx =µₒix²/r²

→B =µₒix/2πr²   ------- (i)


(a) For x =2 cm =0.02 m

  r =10 cm =0.10 m, i =5.0 A

Hence from (i),

B =(4πx10⁻⁷*5*0.02)/(2π*0.10²) T

 =2.0x10⁻⁶ T

 =2.0 µT



(b) For x = r =10 cm =0.10 m

B =µₒi/2πr

  =(4πx10⁻⁷*5)/(2π*0.10) T

 =10x10⁻⁶ T

 =10.0 µT


(c) When the point is outside the wire, i.e. x>r, the current enclosed inside the ampere loop =i. The circulation ∮B.dl over this loop gives,

∮B.dl =µₒi

→B*2πx =µₒi

→B =µₒi/2πx ------- (ii)  

For x =20 cm =0.20 m, 

B =(4πx10⁻⁷*5)/(2π*0.20) T

 =5x10⁻⁶ T

 =5.0 µT.  

Graph of B vs x 

From (i) we see that when the point is inside the wire B=µₒix/2πr² =K*x

Hence B varies directly with x, the graph is linear. 

From (ii) when the point is outside the wire B =µₒi/2πx =K'/x. 

So B is inversely proportional to x. The graph is a non-linear one with B =0 at x =∞. See below    
Graph for Q-50


--------------------------------------------------- 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
Labels: , , , ,

No comments:

Post a Comment

Subscribe to: Post Comments (Atom)