Saturday, August 7, 2021

H C Verma solutions, MAGNETIC FIELD, Chapter-34, Exercises, Q11 -Q20, Concepts of Physics, Part-II

Magnetic Field


Exercises, Q11 to Q20


    11.  A current of 5·0 A exists in the circuit shown in the figure (34-E3). The wire PQ has a length of 50 cm and the magnetic field in which it is immersed has a magnitude of 0·20 T. Find the magnetic force acting on the wire PQ.   
The figure for Q-11


ANSWER: Given, current i =5·0 A, the length of the wire, l =50 cm =0·50 m, the strength of the magnetic field, B =0·20 T. 

   Since the directions of current and the magnetic field are at right angles, the magnetic force on the wire PQ, 

   F =ilB  

     =5·0*0·50*0·20 N

     =0·50 N, towards the battery in the plane of paper/circuit (From the right-hand rule).         





    12.  A circular loop of radius a, carrying a current i, is placed in a two-dimensional magnetic field. The center of the loop coincides with the center of the field (figure 34-E4). The strength of the magnetic field at the periphery of the loop is B. Find the magnetic force on the wire.    
The figure for Q-12


ANSWER: The magnetic field B is perpendicular to the wire and hence current, everywhere. So the magnetic force on the wire,   

F = ilB 

   = i*2πa*B 

   =2πaiB  

 Its direction will be perpendicular to the plane of the loop and going into it.             



 

 

    13.  A hypothetical magnetic field existing in a region is given by B =Bₒeᵣ, where eᵣ denotes the unit vector along the radial direction. A circular loop of radius a, carrying a current i, is placed with its plane parallel to the X-Y plane and the center (0, 0, d). Find the magnitude of the magnetic force acting on the loop.    


ANSWER: Let us first draw a diagram to understand the situation as below:-
The diagram for Q-13


C is the center of the loop and A is a point on the loop in the YZ plane (for example). The direction of the unit vector eᵣ is radial along OA. ∠COA =θ (say). We resolve the magnetic field B in two components along perpendicular directions, Bz perpendicular to the plane of the loop and By along the radius of the loop. The direction of the magnetic force on a very small length dl of the loop due to Bz will be towards the center of the loop and hence the net force on the whole loop due to the components of the magnetic field along the Z-axis will be zero. The direction of the magnetic force on dl due to the component By will be parallel to the Z-axis. The total force on the loop due to By is

F =ilxBy  

F =|F| =ilBy 

   =i*(2πa)*Bₒ.sinθ  

   =2πaiBₒ*{a/√(a²+d²)}

   =2πa²iBₒ/√(a²+d²) 

along either of the Z-axis directions depending upon whether the current in the loop is clockwise or anticlockwise.      





 

    14.  A rectangular wire loop of width 'a' is suspended from the insulated pan of a spring balance as shown in figure (34-E5). A current 'i' exists in the anticlockwise direction in the loop. A magnetic field B exists in the lower region. Find the change in the tension of the spring if the current in the loop is reversed.     
The figure for Q-14



ANSWER: The magnetic forces on the two vertical lengths of the rectangular loop will be horizontal and equal and opposite, so they will not contribute to the tension of the spring. The direction of the magnetic force on the lower width of the loop will be upwards and equal to =iaB. 

  When the current in the loop is reversed, the directions of the horizontal forces on the two lengths of the loop are also reversed but they are still equal and opposite canceling each other. Thus they have still no effect on the tension of the spring. The direction of the magnetic force on the lower width is also reversed and it is downwards with the same magnitude. So now force on it =-iaB. 

   Change in the spring tension 

=iaB-(-iaB) =2iaB.        





 

    15.  A current loop of arbitrary shape lies in a uniform magnetic field B. Show that the net magnetic force acting on the loop is zero.   


ANSWER: Let us consider a very small length of the arbitrary loop at a point. The magnetic force dF =idlxB. Here dl and B are vectors. The total force on the loop,

F =∫dF =∫idlxB = i*∫dlxB 

Since dl is a vector, hence its integration or summation between two points will be a vector joining directly that two points. Since dl is being integrated over a loop here, the two points here are the same. So the distance between these two points is zero, thus the magnitude of ∫dl =0. So the total magnetic force on the loop,

F = i*∫dlxB =0. 

 

  

 



 

    16.  Prove that the force on a current-carrying wire, joining two fixed points a and b in a uniform magnetic field is independent of the shape of the wire.     


ANSWER: Consider an infinitesimally small section of the wire dL. In a uniform magnetic field B, the small force on the wire dL is,  

dF = i*dLxB 
The diagram for Q-16

The total force on the wire 

F = i*∫dLxB   ------------- (i)

dL is a vector hence ∫dL is the sum of all vectors having magnitude dL. All these small vectors are arranged touching tip and toe between points a and b. Hence their vector summation will be a vector having direction ab and magnitude of straight length ab.  Thus ∫dL = L. The total force is now from (i),

F = i*∫dLxB = i*LxB. 

So the magnetic force is always as if on a straight length of wire ab, whatever be the shape of the wire. 




   


 

    17.  A semicircular wire of radius 5·0 cm carries a current of 5·0 A. A magnetic field B of magnitude 0·50 T exists along the perpendicular to the plane of the wire. Find the magnitude of the magnetic force acting on the wire.    


ANSWER: The magnetic force on the semicircular wire will be equivalent to the force on a wire length of the diameter of the loop. Hence the magnitude of the magnetic force, 

F =iLB =i*(2r)*B 

  =5·0*(2*5·0/100)*0·5 N 

  =0·25 N.        





 

    18.  A wire carrying a current i, is kept in the X-Y plane along the curve y =A.sin(2πx/λ). A magnetic field B exists in the z-direction. Find the magnitude of the magnetic force on the portion of the wire between x =0 and x =λ.    


ANSWER: The magnetic force on a current-carrying curved wire length between points a and b is the same as the magnetic force on a straight wire length ab. For the curve length in the given problem, the straight length between two points x =0 and x = λ is equal to λ. Hence the magnitude of the magnetic force on the given portion of the current-carrying wire,

F = i*λ*B {Since the direction of B is perpendicular to λ.} 

   =iλB.

 




 

    19.  A rigid wire consists of a semicircular portion of radius R and two straight sections (figure 34-E6). The wire is partially immersed in a perpendicular magnetic field B as shown in the figure. Find the magnetic force on the wire if it carries a current i.   
The figure for Q-19


ANSWER: Whatever be the shape of the current-carrying wire in a magnetic field, the magnitude of the magnetic force is always equal to the force on a straight length of the wire between endpoints. In the given figure the straight length between endpoints, L =2R. The direction of the magnetic field is perpendicular to the plane of wire, hence the magnitude of the magnetic force,

F =iLB

  =i*2R*B

  =2iRB.  

From the right-hand rule, the direction of this force is upward in the plane of wire.    





 

    20.  A straight, horizontal wire of mass 10 mg and length 1·0 m carries a current of 2·0 A. What minimum magnetic field B should be applied in the region so that the magnetic force on the wire may balance its weight?   


ANSWER: The minimum required magnetic field B should be perpendicular to the length of the wire. In this case, the magnitude of the magnetic force on the wire will be, 

F =iLB.  

F should be equal to the weight of the wire to balance it =10 mg  

=1x10⁻⁵ kg force

=1x10⁻⁵*9·8 N

=9·8x10⁻⁵ N. 

i =2·0 A, L =1·0 m,  

Hence B =F/iL 

  =9·8x10⁻⁵/(2·0*1·0) T 

  =4·9x10⁻⁵ T.                    

  

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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

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Click here for → Exercises (11-20)

Click here for → Exercises (21-30)

Click here for → Exercises (31-42)

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CHAPTER- 7 - Circular Motion

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For more practice on problems on friction solve these- "New Questions on Friction".

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CHAPTER- 5 - Newton's Laws of Motion


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Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)

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CHAPTER- 4 - The Forces

The Forces-

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CHAPTER- 3 - Kinematics - Rest and Motion

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CHAPTER- 2 - "Physics and Mathematics"

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