Saturday, October 16, 2021

H C Verma solutions, MAGNETIC FIELD DUE TO A CURRENT, Chapter-35, EXERCISES, Q11_Q20, Concepts of Physics, Part-II

Magnetic Field Due to a Current


EXERCISES, Q11 to Q20


    11.  Two long, straight wires, each carrying a current of 5 A, are placed along the X - and Y - axes respectively. The currents point along the positive directions of the axes. Find the magnetic fields at the points (a) (1 m, 1 m), (b) (-1 m, 1 m), (c) (-1 m, -1 m) and (d) (1 m, -1 m).   


ANSWER: Given, i = 5 A. If the current is coming towards the viewer, the magnetic field direction around the wire is anticlockwise.  
Figure for Q-11

 From this rule, the direction of the magnetic field at A due to the wire along the X-axis is perpendicular to the paper and coming out. The magnitude can be found by putting i =5 A and d =1 m in B =µ₀i/2πd. The magnetic field at A due to the wire along Y-axis is perpendicular to the paper and going in with the same magnitude. So both the fields are equal in magnitude but opposite in direction, thus canceling out. The resultant magnetic field is zero at A.  

   At point B, the distance of the point from the wires and the current in the wires is the same and hence magnitudes are equal. From the rule, the directions of the magnetic fields at B are the same due to each wire and hence they will add up. So, the net magnetic field at point B is,

=2*µ₀i/2πd 

=µ₀i/πd 

=(4πx10⁻⁷)*5/π*1 T

=2.0x10⁻⁶ T  

=2.0 µT, perpendicularly coming out of the XY plane at point B.         

   At point C, the magnetic fields due to each wire are equal in magnitude but opposite in direction, hence the resultant magnetic field at point C (-1 m, -1 m) is zero. 

   At point D, the magnitude of magnetic fields due to each wire is equal and also in the same direction (going into the plane). The resultant magnetic field at D will be, 

=2µₒi/2πd =µₒi/πd 

=(4πx10⁻⁷)*5/π*1 T 

=2.0 µT, perpendicularly going into the XY plane at point D.     




  


    12.  Four long, straight wires, each carrying a current of 5.0 A, are placed in a plane as shown in figure (35-E3). The points of intersection form a square of side 5.0 cm. (a) Find the magnetic field at the center P of the square. (b) Q₁, Q₂, Q₃ and Q₄ are points situated on the diagonals of the square and at a distance from P that is equal to the length of the diagonal of the square. find the magnetic field at these points.   
Figure for Q-12


ANSWER: (a) Distance of center P from each wire is the same and the magnitude of current in each wire is also the same, hence the magnitude of the magnetic field due to each wire is the same. As we see the magnetic fields due to each set of the parallel wire at P is equal but opposite in direction, thus the resultant field at center P is zero. 

     (b) The distance of these points from the nearer wires is equal to r =2.5 cm and from the farther wires is r' =7.5 cm (From geometry). 

   Point Q₁: From the rule, the direction of magnetic fields due to each wire is the same here hence all will add up. The resultant field, 

B =2*µₒi/2πr +2*µₒi/2πr'

 =(µₒi/π)*(1/r +1/r') 

 =(4πx10⁻⁷*5/π)(1/2.5 +1/7.5)*100 T

 =(2x10⁻⁶)(40+13.3) T

 ≈1.1x10⁻⁴ T, coming out of the plane of the figure perpendicularly. 


     Point Q₂: The magnetic field at this point by each set of parallel wires is equal in magnitude but opposite in direction. Hence the net magnetic field is zero. 


   Point Q₃: The direction of magnetic fields at this point due to a set of parallel wires is the same and magnitude equal. Hence they will add up. The location of the point is similar to the point Q₁ hence the same resultant magnitude 1.1x10⁻⁴ T but the direction is perpendicularly going in the plane of the figure.    


   Point Q₄: Its position is similar to point Q₂. The direction of the magnetic field due to the horizontal parallel set of wires is going into the plane of the figure while it is just opposite due to the vertical parallel set of wires. The magnitudes being equal. Hence the net magnetic field is zero. 





  

    13.  Figure (35-E4) shows along wire bent at the middle to form a right angle. Show that the magnitudes of the magnetic fields at the points P, Q, R and S are equal and find this magnitude.  
Figure for Q-13


ANSWER: The magnitude of the magnetic field at the point P due to the vertical wire is zero because it lies inside and on the axis of the wire. Due to the semi-infinite horizontal wire the magnitude of the magnetic field =µₒi/4πd. 

   Hence the net magnitude of the magnetic field at point P =µₒi/4πd. 

     The same is the case with point R that lies on the axis of the vertical wire and the field is only due to the horizontal wire. So the net magnitude of the magnetic field at point R is only due to the horizontal semi-infinite wire =µₒi/4πd. 

   Now point Q and S lie on the axis of the horizontal wire hence the magnetic fields due to this wire at these points are zero. The magnetic fields at these points are only due to the semi-infinite vertical wire. Distance of the points from the vertical wire =d. Hence net magnetic field at each of these points =µₒi/4πd.           




 

    14.  Consider a straight piece of length x of a wire carrying a current i. Let P be a point on the perpendicular bisector of the piece, situated at a distance d from its middle point. Show that for d>>x, the magnetic field at P varies as 1/d² whereas for d<<x, it varies as 1/d.   


ANSWER: The magnitude of the magnetic field at a point on the perpendicular bisector of a current-carrying wire of length x is, 

  B =µₒix/{2πd√(x²+4d²)}

When d>>x, then x² is negligible in compaision to d², hence for the expression x²+4d² we can write only 4d². so now,   

B =µₒix/{2πd√(4d²)} 

  =µₒix/{2πd*2d} 

  =µₒix/{4πd²} 

→B ∝ 1/d². {except d, rest are constant at the given time}


For d<<x, we can ignore 4d² in comparision to x². Hence the magnetic field now,

  B =µₒix/{2πd(√x²)}

    =µₒix/(2πdx)

    =µₒi/(2πd)

Hence, B ∝ 1/d.

 




 

    15.  Consider a 10 cm long piece of a wire which carries a current of 10 A. Find the magnitude of the magnetic field due to the piece at a point that makes an equilateral triangle with ends of the piece.  


ANSWER: Clearly the point lies on the perpendicular bisector of the length of the wire. Here the length of the wire 'a' =10 cm =0.10 m, current in the wire i =10 A and the distance of the point from the wire d =a*sin60° =√3a/2. See diagram below.     
Diagram for Q-15

    Now the magnetic field at the given point, 

B =µₒia/{2πd√(a²+4d²)} 

 =µₒia/{2πd√(a²+4*3a²/4)

 =µₒia/{2πd√(4a²)}

 =µₒia/(4πad)

 =µₒi/(4πd)

 =4π*10⁻⁷*10/(4π*√3*0.10/2) T

 =2x10⁻⁵/√3 T 

 =1.15x10⁻⁵ T

 =11.5x10⁻⁶ T

 =11.5 µT.    




 

    16.  A long, straight wire carries a current i. Let B₁ be the magnetic field at a point P at a distance d from the wire. Consider a section of length l of this such that the point P lies on a perpendicular bisector of the section. Let B₂ be the magnetic field at this point due to this section only. Find the value of d/l so that B₂ differs from B₁ by 1%.   


ANSWER: B₁ =µₒi/2πd. The magnetic field due to length l of the wire at distance d on its perpendicular bisector is,  

B₂ =µₒil/{2πd√(l²+4d²)} 

For the given condition,        


B₂ =0.99*B₁

→B₂/B₁ =0.99
→l/√(l²+4d²) =0.99 
→l²/(l²+4d²) =0.99²
→(l²+4d²)/l² =1/0.99²
→1 +4(d/l)² =1/0.99²
→(d/l)² =¼{1/0.99² -1} =¼(0.02)
→d/l =√{0.005}
     =0.07 

 

 

 

 

 

    17.  Figure (35-E5) shows a square loop ABCD with edge length a. The resistance of the wire ABC is r and that of ADC is 2r. Find the magnetic field B at the center of the loop assuming uniform wires. 
Figure for Q-17


ANSWER: Distance of the center of the loop from each side of the square is, 

d =a/2.  

      Potential difference between points A and C will be the same for each wire ABC and ADC. Let current in ABC =i₁ and in ADC =i₂. Hence, 

  i₁*r =i₂*2r 

→i₁ =2i₂. 

But i =i₁ +i₂ =3i₂ 

→i₂ =i/3,

and i₁ =2i/3 

   The magnetic field due to AB and BC at the center will be the same in magnitude and direction. So total magnetic field at center due to AB and BC, 

B'=2*µₒi₁a/{2πd√(a²+4d²)} 

(Since the center is on the perpendicular bisector of each side) 

=µₒ(2i/3)a/{π(a/2)√(a²+a²)}

=4µₒi/{3πa√2} 

=2√2µₒi/(3πa). Going into the paper.


    Similarly, the magnetic fields due to the sides AD and DC at the center will be equal in magnitude and direction. So the magnitude of the magnetic field due to AD and DC at the center is,

B"=2*µₒi₂a/{2πd√(a²+4d²)}

=µₒ(i/3)a/{π(a/2)√(a²+a²)}

=2µₒi/(3πa√2)  

=√2µₒi/(3πa). Coming out of the paper. 

  So, the net magnetic field due to the loop will be =B' -B" 
=2√2µₒi/(3πa) -√2µₒi/(3πa)  
=√2µₒi/(3πa), going into the paper.   

              




 

    18.  Figure (35-E5) shows a square loop of edge 'a' made of a uniform wire. A current i enters the loop at point A and leaves it at point C. Find the magnetic field at point P which is on the perpendicular bisector of AB at a distance a/4 from it.  
Figure for Q-18


ANSWER: It is clear from the figure that the current on each side of the loop is i/2. Given point, P is at the middle (equidistant) of two parallel current-carrying equal wire segments AD and BC. Since equal currents flow in these two parallel wires hence the magnetic fields due to each at P is equal in magnitude but opposite in directions. Thus net magnetic field at P due to AD and BC is zero. 

   Due to wire AB, the magnetic field at P is,  

B =µₒ(i/2)a/{2π(a/4)√(a²+4a²/16)} 

 =µₒi/{πa√5/2}

=2µₒi/(√5πa), coming out of the paper.

   The magnetic field due to the side DC at point P is, 

B' =µₒ(i/2)a/{2π(3a/4)√(a²+4*9a²/16)}

=µₒi/{3πa√13/2}

=2µₒi/(3√13πa), going into the paper.

  Hence the net magnetic field at point P,

=B -B'

=2µₒi/(√5πa) -2µₒi/(3√13πa)

=(2µₒi/πa)*{1/√5 -1/(3√13)},

Coming out of the paper.       






 

    19.  Consider the situation described in the previous problem. Suppose the current i enters the loop at point A and leaves it at point B. Find the magnetic field at the center of the loop.  


ANSWER: Let the resistance of each side of the wire loop =r. Then the resistance of wire ADCB =3r and that of AB =r. Let current in ADCB =i₁ and in AB =i₂. Equating the potential difference between A and B, we get  

   i₁*3r =i₂*r 

→i₁ =i₂/3.

But i =i₁ +i₂ =i₂/3 +i₂

→i =4i₂/3

→i₂ =¾i 

and i₁ =¼i.

   The center point of the loop is equidistant from each side, d =a/2 and also at the perpendicular bisectors. The magnitudes and the directions of the magnetic fields due to each of the side AD, DC and CB is at the center is the same. Hence total magnetic field at the center due to these three sides, 

B =3*µₒ(¼i)a/{2π(a/2)√(a²+4a²/4)} 

 =3µₒi/(4√2πa), going into the paper.


  The magnetic field at the center due to the wire AB is,

B' =µₒ(¾i)a/{2π(a/2)√(a²+4a²/4)}

  =3µₒi/(4√2πa), coming out of the paper.

  Hence the net magnetic field at the center =B -B'

  =3µₒi/(4√2πa) -3µₒi/(4√2πa)

  =Zero.   





 

    20.  The wire ABC shown in figure (35-E7) forms an equilateral triangle. Find the magnetic field B at the center O of the triangle assuming the wires to be uniform.  
Figure for Q-20


ANSWER: The current coming at A will divide equally into the two sections of the loop. From symmetry the magnetic field due to sides AB and AC at O will be equal and opposite. Hence the total magnetic field due to AB and AC at O is zero.     

   Similarly, the length of each base half is equal, the direction of current in them is equal but opposite. From symmetry the magnetic field due to each half at O is equal but from the rule the direction of them is opposite. Thus the total magnetic field due to each half of the base wire is zero at O. 

    So the resultant magnetic field at center O of the triangle is zero.        

  

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