Saturday, September 4, 2021

H C Verma solutions, MAGNETIC FIELD, Chapter-34, Exercises, Q51 -Q61, Concepts of Physics, Part-II

Magnetic Field


Exercises, Q51 to Q61


    51.  A particle having mass m and charge q is released from the origin in a region in which electric field and magnetic field are given by  

B =-Bₒj and E =Eₒk

Find the speed of the particle as a function of its z-coordinate. 



ANSWER: Along the Z-axis, only the electric force is acting on the particle =qEₒ 

So the acceleration along the Z-axis,  

a =qEₒ/m 

So from the v² = u² +2as, the speed v of the particle after a distance z, 

v² =0² +2(qEₒ/m)z 

→v =√(2qEₒz/m).             






 

    52.  An electron is emitted with negligible speed from the negative plate of a parallel plate capacitor charged to a potential difference of V. The separation between the plates is d and a magnetic field B exists in the space as shown in figure (34-E20). Show that the electron will fail to strike the upper plate if

d > {2mₑV/eBₒ²}½.   
The figure for Q-52


ANSWER: Electric field between the plates, E =V/d. 

  Coulomb Force on an electron =eE

  =eV/d

Acceleration due to this force = eV/mₑd

Speed after a distance x, from v² = u²+2as; 

v² =0 +2eVx/mₑd 

→v =√(2eVx/mₑd) 

Radius of the circle due to this speed, 

=mₑv/eB 

=(mₑ/eB)*√(2qVx/mₑd) 

=√(2mₑVx/edB²) 

The maximum distance it can travel towards the upper plate is the radius of this circle. For the x =d, 

Radius =√(2mₑVd/edB²) 

  =√(2mₑV/eB²) 

So if d >√(2mₑV/eB²), the electron will fail to strike the upper plate.         





    53.  A rectangular coil of 100 turns has a length of 5 cm and a width of 4 cm. It is placed with its plane parallel to a uniform magnetic field and a current of 2 A is sent through the coil. Find the magnitude of the magnetic field B, if the torque acting on the coil is 0·2 N-m.   


ANSWER: Area of the coil A =5*4 cm²

→A =20 cm² =0.002 m²,  

i =2 A, Number of turns n =100.

   The directions of the area vector and the magnetic field are mutually perpendicular. Hence, 

The torque =n*i*A*B 

  =100*2*0.002*B 

  =0.4B

But given this torque =0.2 N-m

So, 0.4B =0.2 

→B =0·5 T.  

         



 


    54.  A 50 turn circular coil of radius 2·0 cm carrying a current of 5·0 A is rotated in a magnetic field of strength 0·20 T. (a) What is the maximum torque that acts on the coil? (b) In a particular position of the coil, the torque acting on it is half of this maximum. What is the angle between the magnetic field and the plane of the coil?    


ANSWER: Number of turn, n =50, B=0·20 T, i =5·0 A, Radius r =2.0 cm =0.02 m, Hence torque Γ =niAB*sinδ, where δ is the angle between the area vector and the magnetic field.

(a) This torque will be maximum if the value of sinδ is maximum. Maximum value of sinδ will be 1 for δ=90°.

 So maximum torque on the coil, 

Γₘₐₓ=niAB

=50*5*{π*(0.02)²}0·2 N-m

≈0·063 N-m

=6·3x10⁻² N-m.     


(b) Now given that for a certain angle, 

Γ =½Γₘₐₓ 

so, niABsinδ =Γ 

→50*5*{π(0.02)²}*0.2*sinδ =½*0.063

→0.04π sinδ =0.063

→sinδ =0.5

→δ =30° 

It is the angle between area vector (perpendicular to the plane of the coil) and the direction of magnetic field. Hence the angle betwen the plane of coil and the magnetic field =90°-30° =60°.   





 

    55.  A rectangular loop of sides 20 cm and 10 cm carries a current of 5·0 A. A uniform magnetic field of magnitude 0·20 T exists parallel to the larger side of the loop. (a) What is the force acting on the loop? (b) What is the torque acting on the loop?   


ANSWER: (a) Since the magnetic field is parallel to the larger sides, there will be no force on the larger sides. The shorter sides will be perpendicular to the magnetic field hence the force on them will be perpendicular to the plane of the loop but opposite in direction because the direction of current is opposite in two shorter sides. 

  Since the direction of forces on the loop are equal in magnitude but opposite in direction, net force on the loop is zero.


(b) The forces acting on the loop are equal in magnitude and opposite in direction but their line of action is not the same. So these two forces make a couple and apply a torque on the loop. The direction of magnetic field and the area vector are perpendicular to each other here. Hence the torque acting on the loop =iAxB  

Its magnitude =iAB

   =5*(0.20*0.10)*0.2 N-m 

  =0·02 N-m.        

Its direction will be perpendicular to both A and B i.e. parallel to the shorter side. 





 

    56.  A circular coil of radius 2.0 cm has 500 turns in it and carries a current of 1.0 A. Its axis makes an angle of 30° with the uniform magnetic field of magnitude 0·40 T that exists in the space. Find the torque acting on the coil.   


ANSWER: Radius of the coil r =2·0 cm =0·02 m. 

Number of turns n =500,   

Current i =1·0 A,  

B =0·40 T 

δ =30° 

The torque acting on the coil,

=niAB.sinδ 

=500*1.0*π(0.02)²*0.40*sin30° 

≈0·13 N-m.           





 

    57.  A circular loop carrying a current 'i' has a wire of total length L. A uniform magnetic field B exists parallel to the plane of the loop. (a) Find the torque on the loop. (b) If the same length of wire is used to form a square loop, what would be the torque? Which is larger?    


ANSWER: (a) If the radius of the circular loop = r, then 

2πr =L

→r =L/2π

Area of the loop A =πr²

→A =π(L/2π)² =L²/4π

Hence the torque on the loop,

Γ =iAB

  =i(L²/4π)B 

  =iL²B/(4π).  


(b) For a squar loop of same wire length, the length of ane side =L/4. 

So area of the square =L²/16  

So torque now is 

Γ' = iAB

   = i(L²/16)B 

   =iL²B/16.    


 Let us find the ratio Γ/Γ'.

Γ/Γ' =16/4π =1.27

So the torque in the first case is nearly 1·27 times larger.     





 

    58.  A square coil of edge l having n turns carries a current i. It is kept on a smooth horizontal plate. A uniform magnetic field B exists in a direction parallel to an edge. The total mass of the coil is M. What should be the minimum value of B for which the coil will start tipping over?   


ANSWER: The torque on the coil due to magnetic field Γ =niAB =nil²B. 

  The weight of the coil will act at the center of the coil and the torque due to the weight will be the restoring one. 

  This torque Γ' =Mg*l/2 =Mgl/2. 
Diagram for Q-58

  At the point of tipping,

Γ =Γ' 

nil²B =Mgl/2 

→B =Mg/2nil.        




 

  

    59.  Consider a nonconducting ring of radius r and mass m which has a total charge q distributed uniformly on it. The ring is rotated about its axis with an angular speed ⍵. (a) Find the equivalent electric current in the ring. (b) Find the magnetic moment µ of the ring. (c) Show that µ =(q/2m)l where l is the angular momentum of the ring about its axis of rotation.   


ANSWER: (a) The charge on the ring passes at a point in one time period T. So the equivalent electric current in the ring i =q/T.

→i =q/(2π/⍵) =q⍵/2π.


(b) Magnetic moment of the ring,

µ =i*A

  =(q⍵/2π)*πr²

  =½q⍵r².


(c) Angular momentum of the ring about its axis of rotation,

l =I⍵, {Where moment of inertia,I=mr²}

 =mr²⍵ 

→⍵ =l/mr²

  Putting this in the expression for magnetic moment, 

µ =½q⍵r²

  =½q*(l/mr²)*r²

  =(q/2m)l. 


 



 


    60.  Consider a nonconducting plate of radius r and mass m which has a charge q distributed uniformly over it. The plate is rotated about its axis with an angular speed ⍵. Show that the magnetic moment µ and the angular momentum l of the plate are related as µ =(q/2m)l.    


ANSWER: Time period of rotation, 

T =2π/⍵

   Equivalent current in a ring on the plate dx width at a distance x from the center, 

 di =charge/time period of rotation

  ={(q/πr²)*2πxdx}/(2π/⍵)

  =(q⍵/πr²)*xdx

Magnetic moment of this ring in the circular plate =current*area 

dµ = (q⍵/πr²)xdx*πx²

  =(q⍵/r²)*x³dx  

  Magnetic moment of the plate,

µ =(q⍵/r²)∫x³dx

  =(q⍵/r²)[x⁴/4]

 Putting the limit of integration x =0 to x =r, we get

µ =(q⍵/r²)*r⁴/4

  =¼q⍵r²  

      Angular momentum of the plate,

l =I⍵

 =(½mr²)⍵

→⍵ =2l/mr²

Eliminating ⍵ from the expression of µ,

µ =¼qr²*(2l/mr²)

  =(q/2m)l.

      





    61.  Consider a solid sphere of radius r and mass m which has a charge q distributed uniformly over its volume. The sphere is rotated about a diameter with an angular speed ⍵. Show that the magnetic moment µ and the angular momentum l of the sphere are related as µ =(q/2m)l.  


ANSWER: Charge density in th sphere,

=q/(4πr³/3) =3q/4πr³. 

  Let us consider a very small surface inside the sphere at distance x from the center having an area =xdθ.dx

(See the diagram below)
Diagram for Q-61

When the sphere is rotated about the diameter XY, this small area rotates in a ring having radius x.sinθ and its center on the diameter XY with an angular speed ⍵.

Equivalent current in this ring,

di =(3q/4πr³)*2πx.sinθ*xdθdx/(2π/⍵)

 =(3q⍵/4πr³)x²sinθ.dθ.dx       

Magnetic moment of this ring, 

dµ =di*π(xsinθ)² 

 =(3q⍵/4πr³)x²sinθ*πx²sin²θdθdx

 =(3q⍵/4r³)x⁴sin³θdθdx

We double integrate it between limits x =0 to r and θ =0 to π.

µ =(3q⍵/4r³)∬x⁴sin³θdθdx

 =(3q⍵/4r³)(r⁵/5)∫sin³θdθ

 =(3q⍵r²/20)[¼(-3cosθ+⅓cos3θ)]

=(3q⍵r²/80)[3-⅓+3-⅓]

=(q⍵r²/5)

The angular momemtum of the sphere 

l =I⍵

→l = 2mr²⍵/5

→⍵ =5l/(2mr²)

Replacing ⍵ in the expression of µ,

µ =(qr²/5)*(5l/2mr²)

→µ =(q/2m)l. 

Hence proved.


Note: Integration of sin³xdx

Sin3x = 3sinx -4sin³x

→sin³x =¼(3sinx -sin3x)

→∫sin³xdx =¼∫(3sinx -sin3x)

  =¼[-3cosx +⅓cos3x]+c 


 

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

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
Labels: , , , ,

No comments:

Post a Comment

Subscribe to: Post Comments (Atom)