Tuesday, September 28, 2021

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

Magnetic Field Due to a Current


OBJECTIVE - II


    1.  The magnetic field at the origin due to a current element idl placed at a position r is  

(a) (µₒi/4π)(dlxr/r³)

(b) -(µₒi/4π)(rxdl/r³)

(c) (µₒi/4π)(rxdl/r³)

(d) -(µₒi/4π)(dlxr/r³)         


ANSWER: (c), (d).    


Explanation:  When the current element is at origin, the magnetic field due to this at a point having position vector r is  

dB =(µₒi/4π)(dlxr/r³)

  Here the current element is at the position vector r, the magnetic field at the origin will be 

  dB =(µₒi/4π)(rxdl/r³)  

  Option (c) is correct.


Also the cross-product, 

 rxdl = -dlxr

Hence also, dB =-(µₒi/4π)(dlxr/r³) 

So option (d) is also correct.                





    2.  Consider three quantities x=E/B, y =√(1/µₒεₒ) and z =l/CR. Here l is the length of a wire, C is a capacitance and R is resistance. All other symbols have standard meanings.  

(a) x, y have the same dimensions.

(b) y, z have the same dimensions.

(c) z, x have the same dimensions.

(d) None of the three pairs have the same dimensions.    


ANSWER: (a), (b), (c).    


Explanation:  x =E/B  

 qE has a dimension of force 

  qvB has also the dimension of force. 

Hence [qE] = [qvB] 

→[qE]/[qB] =[v] 

→[E/B] = [v]

→[x] = dimensions of velocity. 


Now, y =√(1/µₒεₒ) 

→y =√(εₒc²/εₒ),  Since µₒ=1/εₒc² 

→y =√c² =c =speed of light 

   =dimensions of velocity.


The third one is, z =l/CR 

 CR =(Q/V)R =QR/V =Q/(V/R) 

→CR =Q/I =Q/(Q/t) =t 

So z =l/t =v 

So z has dimensions of velocity.

Thus options (a), (b) and (c) are correct.                




    3.  A long straight wire carries a current along Z-axis. One can find two points in the X-Y plane such that

(a) the magnetic fields are equal  

(b) the directions of the magnetic fields are the same

(c) the magnitudes of the magnetic fields are equal

(d) the field at one point is opposite to that at the other point.        


ANSWER: (b), (c), (d).    


Explanation:  In the given situation the direction of the magnetic field at any point in the X-Y plane will be perpendicular to the line joining the point and Z-axis and in the X-Y plane. So the lines having equal magnitudes of the magnetic field will be concentric circles having centers at Z-axis in the X-Y plane. Option (c) is correct. The direction of the field at any point on such circles will be along the tangent at that point. So no two points in the X-Y plane will have the same magnitude and direction. Option (a) is not correct. 

  At any point on a radial line from the origin in the X-Y plane, the direction of the magnetic field will be the same. Option (b) is correct. 

   At the diametrically opposite points on the equal magnetic field circles, the fields are opposite. Option (d) is correct.        




    4.  A long straight wire of radius  R carries a current distributed uniformly over its cross-section. The magnitude of the magnetic field is 

(a) maximum at the axis of the wire

(b) minimum at the axis of the wire

(c) maximum at the surface of the wire

(d) minimum at the surface of the wire.      


ANSWER: (b), (c).    


Explanation:  The magnitude of the magnetic field, 

B =µₒi/2πd

So maximum magnitude of the magnetic field will be at the surface of the wire. Option (c) is correct. 

   At the axis, the direction of the magnetic fields due to the current around it will be equal in every direction, resulting in the net-zero magnitude of the field. Hence the option (b) is correct. 


     




    5.  A hollow tube is carrying an electric current along its length distributed uniformly over its surface. the magnetic field 

(a) increases linearly from the axis to the surface

(b) is constant inside the tube

(c) is zero at the axis

(d) is zero just outside the tube?     


ANSWER: (b), (c).    


Explanation:  Like the gravitational field inside a shell, the magnetic field inside a hollow tube is constant. And it will be zero at the axis. Hence the options (b) and (c) are correct. 



  




    6.  In a coaxial straight cable the central conductor and the outer conductor carry equal currents in opposite directions. The magnetic field is zero  

(a) outside the cable

(b) inside the inner conductor

(c) inside the outer conductor

(d) in between the two conductors.        


ANSWER: (a).    


Explanation:  Consider a circle of radius r outside the cable as an Ampereian loop. Then from Ampere's law, 

∮B.dl =µₒi

→B∮dl =µₒ(i-i)

→B*2πr =0

→B =0.  

So outside the cable, the magnetic field is zero. Option (a) is correct. 

  Adjusting the value of r according to the three other options, we can see that the sum of the currents enclosed by the circle is not zero. Hence other three options are not correct.        




    7.  A steady electric current is flowing through a cylindrical conductor.

(a) The electric field at the axis of the conductor is zero.

(b) The magnetic field at the axis of the conductor is zero.

(c) The electric field in the vicinity of the conductor is zero. 

(d) The magnetic field in the vicinity of the conductor is zero.        


ANSWER: (b), (c).    


Explanation:  There is an electric field inside the conductor created by a battery or other source due to which the free electrons flow as current. Option (a) is not correct.  

   The magnetic field in the vicinity of the conductor =µₒi/2πd, not zero. Option (d) is incorrect. 

   The magnetic field at the axis due to all current elements will have equal magnitudes and direction all around it in a cross-section, thus canceling each other. The net magnetic field is zero. Option (b) is correct.  

   Since a current-carrying conductor is electrically neutral, the electric field in its vicinity is zero from the Gauss law. Option (c) is correct.    



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CHAPTER- 28- Heat Transfer

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CHAPTER- 24-Kinetic Theory of Gases







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CHAPTER- 17 - Light Waves




CHAPTER- 14 - Fluid Mechanics



CHAPTER- 13 - Fluid Mechanics


CHAPTER- 12 - Simple Harmonic Motion








CHAPTER- 11 - Gravitation




CHAPTER- 10 - Rotational Mechanics




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CHAPTER- 8 - Work and Energy

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CHAPTER- 3 - Kinematics - Rest and Motion

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