Friday, October 8, 2021

H C Verma solutions, MAGNETIC FIELD DUE TO A CURRENT, Chapter-35, EXERCISES, Q1_Q10, Concepts of Physics, Part-II

Magnetic Field Due to a Current


EXERCISES, Q1 to Q10


    1.  Using the formula F = qvxB and B =µₒi/2πr, show that the SI units of the magnetic field B and the permeability constant µₒ may be written a N/A-m and N/A² respectively.    


ANSWER: F =qvBsinθ 

Sinθ is a ratio, hence no unit. Now,

B = F/qv

   =N/C*(m/s)

   =N-s/(A-s)m

   =N/A-m

 Coulomb, C has been replaced by ampere-second A-s. 


B =μₒi/2πr

μₒ =2πrB/i

    = m(N/A-m)/A

   = N/A²

2π is unitless.   





    2.  A current of 10 A is established in a long wire along the positive Z-axis. Find the magnetic field B at the point (1 m, 0, 0).    


ANSWER: If the current is coming towards the viewer, the magnetic field around the wire is anticlockwise. So the direction of the magnetic field at the given point will be parallel to the positive Y-axis.   
Diagram for Q-2

   The magnitude of the magnetic field,

B =μₒi/2πr  

   = (4π*10⁻⁷)*10/(2π*1) T  

   =2x10⁻⁶ T 

   =2 µT, along the positive Y-axis.   



 

    3.  A copper wire of diameter 1.6 mm carries a current of 20 A. Find the maximum magnitude of the magnetic field B due to the current.    


ANSWER: B =µₒi/2πr 

Since B is inversely proportional to r, so B will increase as we go near the wire. It will be maximum at the surface because further reducing r we go inside the wire where the field begins to decrease. 

For maximum B, 

r =1.6/2 mm =0.8 mm =8x10⁻⁴ m 

So, B =(4π*10⁻⁷)*20/(2π*8x10⁻⁴) T 

    =5x10⁻³ T  

    =5 mT.  



 

    4.  A transmission wire carries a current of 100 A. What would be the magnetic field at a point on the road if the wire is 8 m above the road?    


ANSWER: Here, i =100 A, r =8 m. Hence,

The magnetic field at the road, 

B =µₒi/2πr  

  =(4π*10⁻⁷)*100/(2π*8) T

  = (200/8)*10⁻⁷ T 

  =25x10⁻⁷ T

  =2.5 µT.    



 

    5.  A long, straight wire carrying a current of 1.0 A is placed horizontally in a uniform magnetic field B =1.0x10⁻⁵ T pointing vertically upward (figure 35-E1). Find the magnitude of the resultant magnetic field at the points P and Q, both situated at a distance of 2.0 cm from the wire in the same horizontal plane.    
The figure for Q-5


ANSWER: i =1.0 A, r =2.0 cm =0.02 m

Hence the magnitude of the magnetic field, 

B' =µₒi/2πr 

  =(4π*10⁻⁷)*1.0/(2π*0.02) T 

  =1.0x10⁻⁵ T. 

   If the current is coming towards the viewer, the magnetic field around the wire is anticlockwise. So the direction of B' at point P is vertically upward while it is vertically downward at point Q. 

  So resultant magnetic field at P =B+B'

 =1.0x10⁻⁵ +1.0x10⁻⁵ T

 =2.0x10⁻⁵ T 

 =20x10⁻⁶ T

 =20 µT. 


The resultant magnetic field at point Q, 

=B -B' 

=1.0x10⁻⁵ -1.0x10⁻⁵ T

=zero.        



 

    6.  A long, straight wire of radius r carries a current i and is placed horizontally in a uniform magnetic field B pointing vertically upward. The current is uniformly distributed over its cross-section. (a) At what points will the resultant magnetic field have maximum magnitude? (b) What will be the minimum magnitude of the resultant magnetic field?     


ANSWER: (a) The maximum magnitude of the magnetic field due to the current in the wire will be near its surface and it will be, B' =µₒi/2πr where r is the radius of the wire. The resultant magnetic field will be maximum at points where the directions of the existing uniform magnetic field and the magnetic field due to the current are the same. If we see along the current, such points will be at the leftmost points of the wire and the resultant magnetic field will be B+B' 

 =B +µₒi/2πr. 


(b) Minimum magnitude of the resultant magnetic field will be at points where the directions of B and B' are opposite i.e.

=|B-B'| 

For the condition that the resultant field is zero, B =B'   

→B =µₒi/2πd

→d =µₒi/2πB

If the radius of the wire is equal to d, r =d, then the resultant magnetic field will be zero at the rightmost points on the surface of the wire while looking along the current. If r<d then the resultant is zero (minimum numerical value) at distance d from the wire. 

  If r>d, then the points at distance d will be inside the wire but we have to get points outside the wire. At the surface, the resultant magnetic field will be 

=B-µₒi/2πr and it will be the minimum magnitude because as we go farther from the wire the subtracting quantity µₒi/2πr will get lesser and the resultant magnitude B-µₒi/2πd will increase.           




   


    7.  A long, straight wire carrying a current of 30 A is placed in an external, uniform magnetic field of 4.0x10⁻⁴ T parallel to the current. Find the magnitude of the resultant magnetic field at a point 2.0 cm away from the wire. 


ANSWER: External magnetic field B =4.0x10⁻⁴ T. Current in the wire, i =30 A, distance of the point d =2.0 cm =0.02 m.

  The magnetic field due to the current, B' =µₒi/2πd 

  =(4πx10⁻⁷)*30/(2π*0.02)  

  =3.0x10⁻⁴ T, perpendicular to the current or B. 

Hence the resultant magnitude of the magnetic field =√(B²+B'²)

  =√{(4.0x10⁻⁴)²+(3.0x10⁻⁴)²} 

  =√(5x10⁻⁴)² 

  =5.0x10⁻⁴ T.      





 

    8.  A long vertical wire carrying a current of 10 A in the upward direction is placed in a region where a horizontal magnetic field of magnitude 2.0x10⁻³ T exists from south to north. Find the point where the resultant magnetic field is zero.   


ANSWER: External magnetic field, B=2.0x10⁻³ T, from south to north. To the west of the wire, the direction of the magnetic field due to the current B' will be from north to south. At the point to the west of wire where the magnitude of B' =B, the resultant magnetic field will be zero. Let such point is at distance d from the wire, then  

µₒi/2πd =B 

→d =µₒi/2πB 

  =(4πx10⁻⁷)*10/(2π*2x10⁻³) 

  =1.0x10⁻³ m 

  =1.0 mm, west to the wire.      






 

    9.  Figure (35-E2) shows two parallel wires separated by a distance of 4.0 cm and carrying equal currents of 10 A along opposite directions. Find the magnitude of the magnetic field B at the points A₁, A₂, A₃, A₄.    
The figure for Q-9


ANSWER: Point A₁ 

Distance of the point from the left wire, d =2 cm =0.02 m 

Magnetic field due to left wire,

B =µ₀i/2πd 

  =(4πx10⁻⁷)*10/(2π*0.02) T

  =1.0x10⁻⁴ T, upwards 

Distance of the point from the right wire, d =6 cm =0.06 m 

Magnetic field due to the right wire, 

B' =(4πx10⁻⁷)*10/(2π*0.06)  

 =3.33x10⁻⁵ T, downwards.

Hence the net magnetic field =B-B' 

 =1.0x10⁻⁴ -3.33x10⁻⁵ T 

 =6.67x10⁻⁵ T or 0.67x10⁻⁴ T.  


Point A₂

Here the directions of magnetic fields due to both wires are the same hence they would add up. Distance from the left wire, d =1 cm =0.01 m and distance from the right wire, d' =3 cm =0.03 m. So the net magnetic field at this point.

=µₒi/2πd +µₒi/2πd'

=(µₒi/2π)(1/d +1/d') 

=(4πx10⁻⁷*10/2π)(1/0.01+1/0.03)

=20x10⁻⁷(133.33) T

=2.67x10⁻⁴ T  


Point A₃

This point is the midpoint on the line joining both wires and the directions and the magnitudes of the magnetic fields due to each wire is the same hence they add up. Net magnetic field 

=2µₒi/2πd 

=(4πx10⁻⁷*10)/(π*0.02) T 

=2.0x10⁻⁴ T. 


Point A₄

This point is equidistant from each wire hence the magnitude of the magnetic field due to each wire is the same, say B. The direction of the field due to each wire will be perpendicular to the line joining the wire to the point as shown below:- 
diagram for Q-9

  Clearly the angle between both fields =90°. Net magnetic field, 

B' =√2B 

  =√2µₒi/2πd  

d =√2*2 cm =0.02√2 m 

So, B' =√2µₒi/(2π*0.02√2)

  =µₒi/(0.04π)

 =(4πx10⁻⁷*10)/(0.04π) T 

 =1.0x10⁻⁴ T.                    






  

    10.  Two parallel wires carry equal currents of 10 A along the same direction and are separated by a distance of 2.0 cm. Find the magnetic field at a point that is 2.0 cm away from each of these wires.    


ANSWER: Suppose A and D two parallel wires separated by 2.0 cm carry equal current i =10 A coming towards the viewer. The magnetic field around each wire will be anticlockwise. Since the given point is equidistant from the wires, the magnitude of the magnetic field due to each wire will be equal, say B. The direction of each field will be perpendicular to the line joining the point and the wire in a plane perpendicular to the wires containing the line joining. See the diagram below:- 
Diagram for Q-10

 The given point and the wires make an equilateral triangle in a plane that is perpendicular to the wires, hence each angle is =60°. Angle DCF =90°, so angle ACF =30°. Angle ACE =90°, so angle ECF =60°. Since both fields B are equal in magnitude, their resultant CG will bisect the angle ECF, so angle FCG =30°. Now the angle ACG =30°+30° =60°. Since alternate angles ACG and CAD are equal, the direction of resultant CG is parallel to the line joining the wires AD. The magnitude of the resultant, 

B' =√{B²+B²+2B²cos60°} 

 =√{2B²(1+0.5)}

 =√{3B²}

 =B√3

 =(µₒi/2πd)*√3

 =(4πx10⁻⁷*10√3/(2π*0.02) T

 =1.7x10⁻⁴ T.    



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