Wednesday, December 15, 2021

H C Verma solutions, PERMANENT MAGNETS, Chapter-36, OBJECTIVE-I, Concepts of Physics, Part-II

Permanent Magnets


OBJECTIVE-I


    1.  A circular loop carrying a current is replaced by an equivalent magnetic dipole. A point on the axis of the loop is in 

(a) end-on position

(b) broadside-on position

(c) both

(d) none.  


ANSWER: (a).   



Explanation:  A point on the magnetic axis of a bar magnet is called an "end-on" position.   






    2.  A circular loop carrying a current is replaced by an equivalent magnetic dipole. A point on the loop is in 

(a) end-on position

(b) broadside-on position

(c) both

(d) none.


ANSWER: (b).   



Explanation:  A position on a perpendicular bisector of the bar magnet is called the "broadside-on" position.  





    3.  When a current in a circular loop is equivalently replaced by a magnetic dipole, 

(a) the pole strength m of each pole is fixed 

(b) the distance d between the poles is fixed

(c) the product 'md' is fixed

(d) none of the above.    


ANSWER: (c).   



Explanation:  When a loop is equivalently replaced by a magnetic dipole, that means the dipole moment of both is the same. So the dipole moment of the loop iA = md is fixed.  





    4.  Let r be the distance of a point on the axis of a bar magnet from its center. The magnetic field at such a point is proportional to

(a) 1/r

(b) 1/r² 

(c) 1/r³

(d) none of the above.     


ANSWER: (d).   



Explanation:  The magnetic field at a point at distance r on end-on position is

B =µₒ*2Mr/{4π(r²-l²)²} 

Which is none of the above.    





    5.  Let r be the distance of a point on the axis of a magnetic dipole from its center. The magnetic field at such a point is proportional to  

(a) 1/r

(b) 1/r² 

(c) 1/r³

(d) none of the above.     


ANSWER: (c).   



Explanation:  For a magnetic dipole the magnetic length 2l may be considered very small in comparison to r, i.e. r>>l. Neglecting l in the expression for the magnetic field at the end-on position, 

B =µₒ*2Mr/4πr⁴

  =µₒ*2M/4πr³.

So, B ∝ 1/r³. 





    6.  Two short magnets of equal dipole moments M are fastened perpendicularly at their centers (figure 36-Q1). The magnitude of the magnetic field at a distance 'd' from the center on the bisector of the right angle is 

(a) µₒM/4πd³  

(b) µₒ√2M/4πd³  

(c) µₒ2√2M/4πd³  

(d) µₒ2M/4πd³   
Figure for Q-6

   


ANSWER: (c).   



Explanation:  These short magnets can be considered magnetic dipoles with dipole moment M each. For a point P, at distance r from the dipoles (as shown in the diagram below), the radial strength of the field due to each dipole

B' =µₒ*2Mcos45°/4πr³

  =µₒ2M/(4√2πr³) 

  =µₒM/(2√2πr³) 
Diagram for Q-6

 The strength of the component perpendicular to OP is B". B" is equal in magnitude due to each dipole but opposite in direction. So they cancel each other. Thus only radial components remain in the same direction. Hence the net field at P

B =B'+B' =2B' 

   =2*(µₒM/2√2πr³)

  =µₒM/(√2πr³) 

  =µₒ√2M/2πr³

 =µₒ2√2M/4πr³

(multiplying numerator and denominator by 2)

 




    7.  Magnetic meridian is 

(a) a point 

(b) a line along north-south 

(c) a horizontal plane

(d) a vertical plane.     


ANSWER: (d).   



Explanation:  Magnetic meridian is a plane passing through a point and magnetic North and South poles.  





    8.  A compass needle that is allowed to move in a horizontal plane is taken to a geomagnetic pole. It 

(a) will stay in north-south direction only 

(b) will stay in east-west direction only

(c) will become rigid showing no movement

(d) will stay in any position.    


ANSWER: (d).   



Explanation:  At the magnetic pole the direction of the magnetic field is vertical. But the needle can move only in the horizontal plane. The horizontal component of the magnetic field here is zero. Hence there will be no effect of the magnetic field on the compass needle and it will stay in any position.  





    9.  A dip circle is taken to the geomagnetic equator. The needle is allowed to move in a vertical plane perpendicular to the magnetic meridian. The needle will stay 

(a) in a horizontal direction only

(b) in the vertical direction only 

(c) in any direction except vertical and horizontal 

(d) in any direction, it is released.    


ANSWER: (d).   



Explanation: At the Geomagnetic equator, the direction of the magnetic field is horizontal in the magnetic meridian. So the component of the magnetic field in a plane perpendicular to the magnetic meridian is zero. So the needle of the dip circle that is free to move in this perpendicular plane will stay in any direction it is released.





    10.  Which of the following four graphs may best represent the current-deflection relation in a tangent galvanometer?   
The figure for Q-10


ANSWER: (c).   



Explanation:  In a tangent galvanometer the relation between the current and deflection is given as, 

i =K*tanθ

Where K is a constant.

For very small θ, tanθ varies nearly linearly. As θ approaches π/2, even a small increase in θ increases tanθ by a great amount. And near π/2 tanθ tends to infinity. 

So in the tangent galvanometer, the deflection-current graph will be a straight line initially, and with the increase in deflection, the graph will curve and become asymptotic to the current axis near π/2. Only graph (c) fulfills the condition.      





    11.  A tangent galvanometer is connected directly to an ideal battery. If the number of turns in the coil is doubled, the deflection will

(a) increase 

(b) decrease

(c) remain unchanged

(d) either increase or decrease.   


ANSWER: (c).   



Explanation:  The magnetic field created by the coil at the center is  

B =µₒin/2r 

When the number of turns is kept double, i.e. =2n, the resistance of the coil is also doubled. It results in reducing the current to half i.e. i/2. Now the magnetic field by the coil is 

µₒ(i/2)(2n)/2r =µₒin/2r =B. 

   Thus the magnetic field remains unchanged and so will be the deflection.       





    12.  If the current is doubled, the deflection is also doubled in 

(a) a tangent galvanometer 

(b) a moving coil galvanometer 

(c) both

(d) none.     


ANSWER: (b).   



Explanation:  In a tangent galvanometer the current is proportional to the tangent of the deflection. So doubling the current will not double the deflection. 

  In a moving coil galvanometer, the current is directly proportional to the deflection. Hence when we double the current the deflection will also get doubled. Option (b) is correct.    





    13.  A very long bar magnet is placed with its north pole coinciding with the center of a circular loop carrying an electric current i. The magnetic field due to the magnet at a point on the periphery of the wire is B. The radius of the loop is a. The force on the wire is 

(a) very nearly 2πaiB perpendicular to the plane of the wire 

(b) 2πaiB in the plane of the wire

(c) πaiB along the magnet

(d) zero.    


ANSWER: (a).   



Explanation:  Since the bar magnet is long, the direction of the magnetic field on the periphery due to the north pole placed at the center of the loop will be very nearly radially outward. So the angle between the length element dl and B will be 90°. The force on the length element dl is the cross product, 

dF =i dlxB

   Clearly, the direction of the force is perpendicular to the plane of the loop. Integrating over the whole loop, the magnitude of the force is,  

 F =i(2πaB)*sin90° =2πaiB.   


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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

Click here for → Questions for Short Answer 

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)

CHAPTER- 6 - Friction

Click here for → Questions for Short Answer

Click here for → OBJECTIVE-I

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)

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

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


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

CHAPTER- 3 - Kinematics - Rest and Motion

Click here for "Questions for short Answers"


Click here for "OBJECTIVE-I"


Click here for EXERCISES (Question number 1 to 10)


Click here for EXERCISES (Question number 11 to 20)


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"

Click here for "Questions for Short Answers"


Click here for "OBJECTIVE-II"

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