Thursday, November 25, 2021

H C Verma solutions, MAGNETIC FIELD DUE TO A CURRENT, Chapter-35, EXERCISES, Q51_Q61, Concepts of Physics, Part-II

Magnetic Field Due to a Current


EXERCISES, Q51 to Q61


    51.  Sometimes we show an idealized magnetic field that is uniform in a given region and falls to zero abruptly. One such field is represented in figure (35-E12). Using Ampere's law over the path PQRS, show that such a field is not possible.     
Figure for Q-51


ANSWER: From Ampere's law, the closed integral of B.dl over the loop PQRS should be µₒ times the current through the loop. ∮B.dl over PQRS is the same as ∮B.dl over the length PS. (Because the direction of field and length over PR and RS are perpendicular and the field over QR is zero.) Hence,

 ∮B.dl =B*PS =µₒ*i =µₒ*0 =0

  Because 'i' through the loop is zero.

It gives B = 0 which is not correct. So the magnetic field can not fall to zero abruptly.

 







    52.  Two large metal sheets carry surface currents as shown in figure (35-E13). The current through a strip of width dl is Kdl where K is a constant. Find the magnetic field at a point P, Q, and R.  
Figure for Q-52


ANSWER: From symmetry, the magnitudes of the magnetic fields at P and R will be the same (say B'), and their directions parallel to the plates because the plates are large. The magnetic field at Q due to each plate will be towards the right. So both the magnitudes will be added due to the same direction. Let net magnitude =B. The direction of the magnetic field everywhere will be parallel to plates because the plates are large.
Diagram for Q-52

 We assume an Ampereian loop EFGH through P and R as shown in the figure. The magnetic field is perpendicular to EH and FG. If EF =L, then from ampere's law  

∮B.dl =µₒi
→2B'L =µₒ(KL-KL) =0, Since the currents in the sheets are equal and opposite.

→B' =0.

So the magnetic fields at P and R are zero.

Diagram for Q-52


Take another Ampereian loop ABCD about the upper plate through P and Q as shown in the figure. Here, ∮B.dl =µₒi 

→B'L+BL =µₒ*KL

→BL =µₒKL

→B =µₒK. (towards the right)

 





 

    53.  Consider the situation of the previous problem. A particle having charge q and mass m is projected from the point Q in a direction going into the plane of the diagram. It is found to describe a circle of radius r between the two plates. Find the speed of the charged particle.  


ANSWER: Near point Q, a uniform magnetic field is there B =µₒK. This field and the direction of the speed are perpendiculars hence a magnetic force F will act on the charged particle.  

F =qvB 

This will provide the centripetal force to describe a circle. F =mv²/r. So, 

qvB =mv²/r 

→v =qBr/m 

→v =q(µₒK)r/m =µₒKqr/m.           





 

    54.  The magnetic field B inside a long solenoid, carrying a current of 5.00 A, 3.14x10⁻² T. Find the number of turns per unit length of the solenoid.   


ANSWER: B =3.14x10⁻² T, i =5.0 A. Let n = number of turns per unit length. Then for a long solenoid,  

B =µₒni

→n =B/µₒi 

  =3.14x10⁻²/(4πx10⁻⁷*5)

 =5000 turns/m.        





 

    55.  A long solenoid is fabricated by closely winding a wire of radius 0.5 mm over a cylindrical non-magnetic frame so that the successive turns nearly touch each other. What would be the magnetic field B at the center of the solenoid if it carries a current of 5 A?  


ANSWER: The radius of the wire =0.5 mm, hence the thickness of one turn in the solenoid =1.0 mm. The number of turns in 1-meter length (1000 mm) of the solenoid, n =1000.  

   Given current i =5 A. 

The magnetic field at the center of the solenoid, 

B =µₒni 

 =4πx10⁻⁷*1000*5 T 

 =2πx10⁻³ T           





 

    56.  A copper wire having a resistance of 0.01 ohm in each meter is used to wind a 400-turn solenoid of radius 1.0 cm and length 20 cm. Find the emf of a battery which when connected across the solenoid will cause a magnetic field of 1.0x10⁻² T near the center of the solenoid.  


ANSWER: Given B =1.0x10⁻² T. The number of turns =400 in 20 cm length. Hence the number of turns per meter, 

n =400/0.20 =2000. 

We have B =µₒni, 

→i =B/µₒn 

  =1.0x10⁻²/(4πx10⁻⁷*2000) A

 =100/8π A. 

The total length of the wire in the solenoid, 

=400*2π*1.0 cm 

=800π cm =8π m 

The resistance of the wire in solenoid 

R =8π*0.01 Ω

  =0.08π Ω  

Let the emf of the battery =V. 

Then V =iR

   =(100/8π)*0.08π volts 

  = 1 volt.                 





 

    57.  A tightly wound solenoid of radius 'a' and length 'l' has n turns per unit length. It carries an electric current i. Consider a length dx of the solenoid at a distance x from one end. This contains ndx turns and may be approximated as a circular current indx. (a) Write the magnetic field at the center of the solenoid due to this circular current. Integrate this expression under proper limits to find the magnetic field at the center of the solenoid. (b) Verify that if l>>a, the field tends to B =µₒni and if a>>l, the field tends to B =µₒnil/2a. Interpret these results. 


ANSWER: (a) The magnetic field at a point on the axis of a loop having current 'i' and at a distance d from its center is 

B =µₒia²/2(a²+d²)ⁿ, where n =3/2.

For a loop at distance x from one end, the center of the solenoid is at a distance d = l/2-x from the loop. 

For the current 'indx' in the loop,   

dB =µₒa²indx/2{a²+(l/2-x)²}ⁿ 

Integrating between the limits, x =0 to x =l, we have 

B =∫dB =(µₒa²in/2)∫dx/{a²+(l/2-x)²}ⁿ

Let l/2-x =y, then dx =-dy and the integration becomes,

B =-(µₒa²in/2)∫dy/{a²+y²}ⁿ

 =-½(µₒa²in)[{y/a²√(a²+y²)}] 

{The above integration has been derived by substituting y=a.tanδ where n =3/2} 

The limits for y is :- if x =0, y =l/2 and if x =l, y =-l/2. Putting the limits ,  

B=-½µₒin[(-l/2)/√(a²+l²/4)-(l/2)/√(a²+l²/4)]

=½µₒinl/√{(4a²+l²)/4} --(1)

=µₒinl/{l√{(1+(2a/l)²}

=µₒin/√{1+(2a/l)²}.         


(b) When l>>a, (2a/l)²≈0 

In such case,

B =µₒin.  

This is the case of a long solenoid. B is independent of l and a.


When a>>l, 4a²+l² ≈4a² 

From (1),

B =½µₒinl/√{(4a²+l²)/4}

 =µₒinl/√{(4a²+l²)}

=µₒinl/√(4a²)

 =µₒinl/2a.

   This is the case of a short solenoid. Here B depends on both l and a.        





 

    58.  A tightly wound, long solenoid carries a current of 2.00 A. An electron is found to execute a uniform circular motion inside the solenoid with a frequency of 1.00x10⁸ rev/s. Find the number of turns per meter in the solenoid.    


ANSWER: Current i = 2.0 A.  

If the magnetic field inside the solenoid be B, then equating the magnetic force on the electron and the centripetal force,

evB =mv²/r

v =eBr/m.   

Here v is the speed of the electron. Time period of the revolution, 

T =2πr/v 

Hence frequency f =1/T =v/2πr  

f =eBr/{m*2πr}

B =2πmf/e

But for a long solenoid,

B =μₒni

Hence, 

μₒni =2πmf/e

n = 2πf/μₒ(e/m)i

 =2π*10⁸/{4π*10–⁷*1.76x10¹¹*2}

 =1.42x10³

 =1420 turns per m.

              





 

    59.  A tightly wound, long solenoid has n turns per unit length a radius r, and carries a current i. A particle having charge q and mass m is projected from a point on the axis in a direction perpendicular to the axis. What can be the maximum speed for which the particle does not strike the solenoid? 


ANSWER: Since the solenoid is long, the magnetic field inside it 

B = μₒni

Force on a particle having a charge q in this magnetic field,

F =qvB, where v is the speed of the particle. This force is perpendicular to the speed. So it will act as the centripetal force to make the particle describe a circular path. Let the radius of the path =R.

→mv²/R =qvB 

→mv =qRμₒni

v =μₒqRni/m

Maximum speed will be when the radius R is maximum. In order that the particle does not strike the solenoid, R =r/2. It is because of the particle being projected from a point on the axis. The largest circle without striking the solenoid will be between the axis and the side of the solenoid. The diameter of this circle =r, hence the radius, R =r/2. So the maximum speed of the particle 

= μₒq(r/2)ni/m 

= μₒqrni/2m.

           





 

    60.  A tightly wound, long solenoid is kept with its axis parallel to a large metal sheet carrying a surface current. The surface current through a width dl of the sheet is Kdl and the number of turns per unit length of the solenoid is n. The magnetic field near the center of the solenoid is found to be zero. (a) Find the current in the solenoid. (b) If the solenoid is rotated to make its axis perpendicular to the metal sheet, what would be the magnitude of the magnetic field near its center?  


ANSWER: First, let us find out the magnetic field at a point P, a distance d from the large metal sheet. 

Consider the Ampereian loop in the diagram below. P and Q are at a distance d from the plate. Hence the magnitude at these points will be the same =B. 
Diagram for Q-60

From Amperes law,

∮B.dl over the loop =2BL

So, 2BL =µₒi

→2BL =µₒ*KL

→B =µₒK/2. {It is independent of d}  


It can also be derived from the diagram below, 

Diagram for Q-60


The magnetic field at P due to the current Kdl in the sheet at distance l from the root of the perpendicular from P, 

dB =µₒKdl/{2π√(d²+l²)} 

{Since B =µₒi/2πr}

Its component parallel to the plate will be only effective because the perpendicular component will be neutralized by the current equidistant from the point Q. So we have to integrate dBsinδ between the limits δ =0 to π. 

 So the magnetic field at P, 

B =∫dBsinδ 

 =∫µₒKdl.sinδ/{2π√(d²+l²)}

 =(µₒK/2π)∫dl*d/(d²+l²)

 =(µₒK/2πd)∫{d²/(d²+l²)}dl

 =(µₒK/2πd)∫sin²δ.dl ----- (i)

 Since d/l =tanδ, →l =d.cotδ 

→dl =-d.cosec²δdδ

Putting in (i), 

B=(µₒK/2πd)∫sin²δ(-d.cosec²δ.dδ)

 =-(µₒK/2π)∫dδ

 =-(µₒK/2π)[δ], putting limis of δ

 =-(µₒK/2π)[0 -π]

 =µₒK/2. 


(a) The magnetic field at the center of the solenoid due to current 'i' in the solenoid =µₒni. Since the net magnetic field at the center is found to be zero, the magnetic fields due to the sheet and the solenoid must be equal and opposite.

Thus, µₒni =µₒK/2

→i =K/2n.


(b) When the solenoid is kept perpendicular to the metal sheet, two equal magnetic fields, each µₒK/2, are directed perpendicular to each other. Hence the resultant magnetic field,

=√{(µₒK/2)²+(µₒK/2)²}

=√{2(µₒK/2)²}

=√2*µₒK/2

=µₒK/√2.

            





 

    61.  A capacitor of capacitance 100 µF is connected to a battery of 20 volts for a long time and then disconnected from it.  It is now connected across a long solenoid having 4000 turns per meter. It is found that the potential difference across the capacitor drops to 90% of its maximum value in 2.0 seconds. Estimate the average magnetic field produced at the center of the solenoid during this period.   


ANSWER: Q =CV. In 2.0 seconds the remaining potential difference =0.9V, so the charge on the capacitor is now Q' =0.9CV =0.9Q  

Thus in this period of 2.0 s, 0.1Q of charge flows through the solenoid. So average current,  

i =0.1Q/2 =0.05Q =0.05CV 

The average magnetic field at the center of the long solenoid, 

B =µₒni

 =µₒn*0.05CV 

 =0.05*4πx10⁻⁷*4000*100x10⁻⁶*20 T 

 =1.6πx10⁻⁷ T 

 =16πx10⁻⁸ 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

Click here for → Question for Short Answers

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → Exercises (1-10)

Click here for → Exercises (11-20)

Click here for → Exercises (21-30)

Click here for → Exercises (31-42)

Click here for → Exercise(43-54)

CHAPTER- 7 - Circular Motion

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Click here for → EXERCISES (11-20)

Click here for → EXERCISES (21-30)

CHAPTER- 6 - Friction

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Click here for → Friction - OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → Exercises (11-20)

Click here for → EXERCISES (21-31)

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

---------------------------------------------------------------------------------

CHAPTER- 5 - Newton's Laws of Motion


Click here for → QUESTIONS FOR SHORT ANSWER

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

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CHAPTER- 4 - The Forces

The Forces-

"Questions for short Answers"    


Click here for "The Forces" - OBJECTIVE-I


Click here for "The Forces" - OBJECTIVE-II


Click here for "The Forces" - Exercises


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CHAPTER- 3 - Kinematics - Rest and Motion

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Click here for EXERCISES (Question number 21 to 30)


Click here for EXERCISES (Question number 31 to 40)


Click here for EXERCISES (Question number 41 to 52)


CHAPTER- 2 - "Physics and Mathematics"

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