Saturday, August 14, 2021

H C Verma solutions, MAGNETIC FIELD, Chapter-34, Exercises, Q21 -Q30, Concepts of Physics, Part-II

Magnetic Field


Exercises, Q21 to Q30


    21.  Figure (34-E7) shows a rod PQ of length 20·0 cm and mass 200g suspended through a fixed point O by two threads of length 20·0 cm each. A magnetic field of strength 0·500 T exists in the vicinity of the wire PQ as shown in the figure. The wires connecting PQ with the battery are loose and exert no force on PQ. (a) Find the tension in the threads when the switch S is open. (b) A current of 2·0 A is established when switch S is closed. Find the tension in the threads now.     
The figure for Q-21


ANSWER: (a) When the switch S is open, there is no current in the rod PQ. With no current, the threads have to balance the weight of the rod PQ.  

   Weight, W =mg

      = 0.20*9.8 N = 1.96 N  

Let the tension in each of the threads =T  

Balancing the forces in the vertical direction, 

2T .sin 60° =W

→  T = W/√3 

      =1.96/√3 N

      =1.13 N


(b) When the switch S is closed, the current in the rod PQ, i =2.0 A, 

Force due to the magnetic field, 

F = ilB,  

(because the current and the magnetic fields are perpendicular to each other).

→F  =2.0*0.20*0.500 N

  =0.20 N

From the right hand rule, this force is downward. Hence, the total force downward =W+F

      = 1.96+0.20

      = 2.16 N

Hence, the tension in each thread 

 = 2.16/√3 =1.25 N   






    22.  Two metal strips, each of length l, are clamped parallel to each other on a horizontal floor with a separation b between them. A wire of mass m lies on them perpendicularly as shown in figure (34-E8). A vertically upward magnetic field of strength B exists in the space. The metal strips are smooth but the coefficient of friction between the wire and the floor is µ. A current i is established when the switch S is closed at the instant t =0. Discuss the motion of the wire after the switch is closed. How far away from the strips will the wire reach?    
The figure for Q-22


ANSWER: The force on the wire will be towards right and equal to,   F =ibB.  

The acceleration of the wire, 

a =ibB/m

Velocity, v, at the end of the strips, 

v² = u² +2as 

   = 0 +2ibBl/m

→ v =√(2iblB/m)

When the wire moves on the floor, the force of friction on it, F' =µmg

Retardation, a' =F'/m=µg 

Suppose it goes upto a distance =d, then from v² = u² +2as 

Here v =0, s =d, a = -a', and    

u = v  = √(2iblB/m) 

Hence, d  = u²/2a'

 →  d = 2iblB/2mµg 

        = ilbB/µmg 

 




 

    23.  A metal wire PQ of mass 10 g lies at rest on two horizontal metal rails separated by 4·90 cm (figure 34-E9). A vertically downward magnetic field of magnitude 0·800 T exists in the space. The resistance of the circuit is slowly decreased and it is found that when resistance goes below 20·0 Ω, the wire PQ starts sliding on the rails. Find the coefficient of friction.    
The figure for Q-23


ANSWER: m = 10 g =0.010 kg 

Force of friction on the wire when trying to move F' =µmg =µ*0.010*9.8 =0.098µ 
  When the wire just starts to move, the magnetic force on the wire is equal to this force of friction. 
But the magnetic force,F =ilB
 F = (6/20)*(4.90/100)*0.800 N  
    =0.0118 N  
So, 0.098µ =0.0118
  → µ =0.12       

        




 

    24.  A straight wire of length l can slide on two parallel plastic rails kept in a horizontal plane with a separation d. The coefficient of friction between the wire and the rails is µ. If the wire carries current 'i', what minimum magnetic field should exist in the space in order to slide the wire on the rails.    


ANSWER: For the magnetic field, B to be minimum, its direction should be perpendicular to the horizontal plane. 

   The magnitude of the magnetic field, 

F =ilB 

If the mass of the wire =m, its weight =mg. 

When the wire tends to slide, the force of friction, F' =µmg

For the wire to slide on the rails, F =F'

→ilB =µmg

→B = µmg/il     




 

    25.  Figure (34-E10) shows a circular wire loop of radius a, carrying a current i, placed in a perpendicular magnetic field B. (a) Consider a small part dl of the wire. Find the force on this part of the wire exerted by the magnetic field. (b) Find the force of compression in the wire.    
The figure for Q-25


ANSWER: (a) Since the wire is placed in a perpendicular magnetic field, the force on a small part dl of the wire carrying a current i will be equal to   

dF =i*dl*B =idlB 

From the figure, the magnetic field is going inside the plane of the loop, so from the right-hand rule, its direction will be towards the center.     


(b) Let the force of compression in the loop = T.

  Considering on section that divides the loop into two equal halves. On the two ends of a half, the force of compression = T at each ends.  Suppose, the part dl is situated at angle θ from the horizontal axis as shown in the figure below.
Diagram for Q-25

     The dl length subtends an angle dθ at the center. Equating the forces along T,    

2T = ∫dF*cosθ

→2T = ∫idlB*cosθ   

→2T =iB∫cosθ*adθ  =iaB∫cosθdθ  

→2T =iaB[ sinθ]

{The limits of integration is from -π/2 to π/2}   

→2T =iaB [ sin π/2 -sin(-π/2)] =2iaB

→T = iaB   




 

    26.  Suppose the radius of cross-section of the wire used in the previous problem is r. Find the increase in the radius of the loop if the magnetic is switched off. The Young's modulus of the material of the wire is Y.    


ANSWER: The length of the wire =circumference of the loop =2πa,  

Area of the cross-section of the wire, A =πr²

  When compression is induced in the wire, the stress in the wire = T/πr²

Suppose the decrease in the length =dl, 

From the theory of elasticity,

Stress/strain = Y

(T/πr²)/(dl/2πa) = Y

dl =2aT/r²Y

Now the length of the wire

 =2πr -2aT/r²Y

The radius of the loop now       

=r -aT/πr²Y

The radius of the loop decreases by

aT/πr²Y. 

   So the radius will increase by the same length when compression vanishes due to the absence of the magnetic field. Putting the value of T in the expression, the increase in the radius of the loop    

=a*iaB/πr²Y

=ia²B/πr²Y      


  


 

  

    27.  The magnetic field existing in a region is given by 

B = Bₒ{1+x/l}k.  

A square loop of edge l and carrying a current i, is placed with its edges parallel to the X-Y axes. Find the magnitude of the net magnetic force experienced by the loop.   


ANSWER: Let us first draw a diagram. 
Diagram for Q-27
   Each side of the square loop is equal to l. Assume the current in the loop as shown.  

Force on the side AB,  

F₁ =ilB

  =ilB₀{1 +l/l} 

  =2ilB₀    (towards right)

Force on the side CO 

F₃ =ilB₀{1 +0/l} =ilB₀ (towards left)

Force on the side BC

F₂ =ilB₀{1 +(l/2)/l} =3ilB₀/2, upward.

(Since B varies linearly along X-axis, here we have taken an average magnetic force at x =l/2)
Similarly the force on the side OA

F₄ =ilB₀{1 +(l/2)/l} =3ilB₀/2, downward

 The forces along Y-axis F₂ =F₄, are equal and opposite so no net force along the Y-axis.

  Net force along X-axis =F₁ -F₃   

  =2ilB₀ -ilB₀  

  =ilB₀.         




 

    28.  A conducting wire of length l, lying normal to a magnetic field B, moves with a velocity v as shown in the figure (34-E11). (a) Find the average magnetic force on a free electron of the wire. (b) Due to the magnetic force, electrons concentrate at one end resulting in an electric field inside the wire. The redistribution stops when the electric force on the free electrons balances the magnetic force. Find the electric field developed inside the wire when the redistribution stops. (c) What potential difference is developed between the ends of the wire?      
The figure for Q-28


ANSWER: (a) The magnetic force on a particle moving with a velocity v and having a charge q in a magnetic field of strength B is    

  F =qvB, when the directions of v and B are perpendicular to each other.  

   The free electrons in the wire will have an average velocity v in the direction of movement of the wire. Charge on an electron =e. Hence the average magnetic force on a free electron in the wire = evB.     

  

(b) Suppose the electric field developed inside the wire = E.  

Electric force on an electron =eE.  

When the electric and magnetic fields balance each other inside the wire, we have,   

eE =evB 

→E =vB.  


(c) Let the potential difference developed between the ends of the wire = V. 

Since E =dV/dr =V/l, (for a uniform electric field)  

→V =lE = lvB.             




 

    29.  A current is passed through a silver strip of width d and area of cross-section A. The number of free electrons per unit volume is n. (a) Find the drift velocity v of the electrons. (b) If a magnetic field B exists in the region as shown in figure (34-E12), what is the average magnetic force on the free electrons? (c) Due to the magnetic force, the free electrons get accumulated on one side of the conductor along its length. This produces a transverse electric field in the conductor which opposes the magnetic force on the electrons. Find the magnitude of the electric field which will stop further accumulation of electrons. (d) What will be the potential difference developed across the width of the conductor due to the electron accumulation? The appearance of the transverse emf, when a current-carrying wire is placed in a magnetic field, is called the Hall effect.  
The figure for Q-29


ANSWER: (a) Assume the length of the strip = l, Volume of the strip =Al. The total number of free electrons in the strip =Aln. 

  The charge of these free electrons 

 q =Alne. 

The current in the strip, i =q/t 

→i =Alne/t =Ane*(l/t)  

→i =Anev, (v is the drift velocity of the elctron) 

→v =i/(Ane). 


(b) Average magnetic force on free electrons, F = evB 

→F =e*(i/Ane)*B

  =iB/An.            

From the right-hand rule, the direction of this force will be upward in the figure. 


(c) Let this balancing electric field inside the strip = E.  

The electric force on an electron =eE. 

For the given condition,

eE =F 

→eE =iB/An

→E =iB/Ane.  


(d) Let the potential difference developed across the width of the strip = V.   

Since E =dV/dr, here, E =V/d,  

→iB/Ane =V/d 

→V = iBd/Ane.         




  


    30.  A particle having a charge of 2·0x10⁻⁸ C and a mass of 2·0x10⁻¹⁰ g is projected with a speed of 2·0x10³ m/s in a region having a uniform magnetic field of 0·10 T. The velocity is perpendicular to the field. Find the radius of the circle formed by the particle and also the time period.    


ANSWER: For the particle given that,

Charge, q =2·0x10⁻⁸ C and mass, 

m = 2·0x10⁻¹⁰ g =2·0x10⁻¹³ kg, The perpendicular Magnetic field, B =0·10 T, velocity of the particle, v =2·0x10³ m/s.  

  Hence the magnetic force on the particle F =qvB    

    Since this force will always be perpendicular to the direction of velocity, the particle will move in a circle with F as centripetal force.  

Thus F = mv²/r, where r is the radius of this circle. 

→r =mv²/F

   =mv²/qvB    

   =mv/qB

   =2x10⁻¹³*2x10³/(2x10⁻⁸*0·10) m 

  =2·0x10⁻¹ m 

  =0·20 m 

  =20 cm.  

   Let the time period be =T. In this time, the particle travels a distance equal to the circumference of the circle, d =2πr. Hence 

T =d/v =2πr/v 

   =2πmv/vqB 

   =2πm/qB

   =2*3.14*2x10⁻¹³/(2x10⁻⁸*0·10) 

   =6.3x10⁻⁴ s                     

 

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