HC Verma Magnetic Field due to a current Solutions of Que for Short Ans Ch-35 Vol-2

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HC Verma Magnetic Field due to a current Solutions of Que for Short Ans Ch-35 Vol-2 Concept of Physics. Step by Step Solution of Questions for short answer of Ch-35 Magnetic Field due to a current HC Verma Question of Bharti Bhawan Publishers . Visit official Website CISCE for detail information about ISC Board Class-12 Physics.

HC Verma Magnetic Field due to a current Solutions of Que for Short Ans Ch-35 Vol-2

Board ISC and other board
Publications Bharti Bhawan Publishers
Ch-35 Magnetic Field due to a current
Class 12
Vol  2nd
writer H C Verma
Book Name Concept of Physics
Topics Solutions of Question for Short Answer
Page-Number 248

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Que for Short Ans

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Exercise


H C Verma Magnetic Field due to a current Que for Short Ans

 Solutions of Ch-35 Vol-2 Concept of Physics for Class-12

(Page – 248)

Question 1 :-

An electric current flows in a wire from north to south. What will be the direction of the magnetic field due to this wire at a point (a) east of the wire, (b) west of the wire, (c) vertically above the wire and (d) vertically below the wire?

Answer 1 :-
According to the right-hand thumb rule :-
If the thumb of our right hand points in the direction of the current flowing, then the curling of the fingers will show the direction of the magnetic field developed due to it and vice versa. Let us consider the case where an electric current flows north to south in a wire.
According to the right-hand thumb rule :-
(a) for a point in the west of the wire, the magnetic field will enter the plane of paper.
(b) for any point vertically above the wire, the magnetic field will be from right to left.
(c) for any point in the east of the wire, the magnetic field will come out of the plane of paper.
(d) for any point vertically below the wire, the magnetic field will be from left to right.

Question 2 :-

The magnetic field due to a long straight wire has been derived in terms of µ, i and d. Express this in terms of ε0, c, i and d.

Answer 2 :-

The magnetic field due to a long, straight wire is given by :

HC Verma Magnetic Field due to a current Solutions of Que for Short Ans Ch-35 Vol-2 img 1
(In terms of ε0, c, i and d)

Question 3 :-

You are facing a circular wire carrying an electric current. The current is clockwise as seen by you. Is the field at the centre coming towards you or going away from you?

Answer 3 :-

According to the right-hand thumb rule :-

If we curl the fingers of our right hand in the direction of the current flowing, then the thumb will point in the direction of the magnetic field developed due to it and vice versa. Hence, the field at the centre is going away from us.

In Ampere’s law ∮B¯⋅dl¯ = μ0i,  ‘i’ is the total current crossing the area bounded by the closed curve.

The magnetic field B on the left-hand side is the resultant field due to all existing currents.

Question 5 :-

The magnetic field inside a tightly wound, long solenoid is B = µ0 ni. It suggests that the field does not depend on the total length of the solenoid, and hence if we add more loops at the ends of a solenoid the field should not increase. Explain qualitatively why the extra-added loops do not have a considerable effect on the field inside the solenoid.

Answer 5 :-
The magnetic field due to a long solenoid is given as B = µ0ni,
Where, n is the number of loops per unit length.
So, if we add more loops at the ends of the solenoid, there will be an increase in the number of loops and an increase in the length, due to which the ratio n will remain unvaried, thereby leading to not a considerable effect on the field inside the solenoid.

Question 6 :-

A long, straight wire carries a current. Is Ampere’s law valid for a loop that does not enclose the wire, or that encloses the wire but is not circular?

Answer 6 :-

Ampere’s law is valid for a loop that is not circular. However, it should have some charge distribution in the area enclosed so as to have a constant electric field in the region and a non-zero magnetic field. Even if the loop defined does not have its own charge distribution but has electric influence of some other charge distribution, it can have some constant magnetic field ( ∮B¯⋅dl¯ = μ0i enclosed  ).

Question 7 :-

A straight wire carrying an electric current is placed along the axis of a uniformly charged ring. Will there be a magnetic force on the wire if the ring starts rotating about the wire? If yes, in which direction?

Answer 7 :-

The magnetic force on a wire carrying an electric current i is F‾=i.(l‾×B‾),

Where, l is the length of the wire and B is the magnetic field acting on it. If a uniformly charged ring starts rotating around a straight wire, then according to the right-hand thumb rule, the magnetic field due to the ring on the current carrying straight wire placed at its axis will be parallel to it. So, the cross product will be

(l‾×B‾)=0

⇒F‾=0

Hence, no magnetic force will act on the wire.

 

Question 8 :-

Two wires carrying equal currents i each, are placed perpendicular to each other, just avoiding a contact. If one wire is held fixed and the other is free to move under magnetic forces, what kind of motion will result?

Answer 8 :-

The magnetic force on a wire carrying an electric current i is F‾=i.(l‾×B‾) , ​

Where l is the length of the wire and B is the magnetic field acting on it.

Suppose we have one wire in the horizontal direction (fixed) and other wire in the vertical direction (free to move). If the horizontal wire is carrying current from right to left is held fixed perpendicular to the vertical wire, which is free to move, the upper portion of the free wire will tend to move in the left direction and the lower portion of the wire will tend to move in the right direction, according to Fleming’s left hand rule, as the magnetic field acting on the wire due to the fixed wire will point into the plane of paper above the wire and come out of the paper below the horizontal wire and the current will flow in upward direction in the free wire.

Thus, the free wire will tend to become parallel to the fixed wire so as to experience maximum attractive force.

Question 9 :-

Two proton beams going in the same direction repel each other whereas two wires carrying currents in the same direction attract each other. Explain.

Answer 9 :-
Two proton beams going in the same direction repelled each other, as they are like charges and we know that like charges repel each other.
When a charge is in motion then a magnetic field is associated with it. Two wires carrying currents in the same direction produce their fields (acting on each other) in opposite directions so the resulting magnetic force acting on them is attractive.
Due to the magnetic force, these two wires attract each other. But when a charge is at rest then only an electric field is associated with it and no magnetic fields is produced by it. So at rest, it repels a like charge by exerting a electric force on it.
Charge in motion can produce both electric field and magnetic field.
The attractive force between two current carrying wires is due to the magnetic field and repulsive force is due to the electric field.

Question 10 :-

In order to have a current in a long wire, it should be connected to a battery or some such device. Can we obtain the magnetic due to a straight, long wire by using Ampere’s law without mentioning this other part of the circuit?

Answer 10 :-

We can obtain a magnetic field due to a straight, long wire using Ampere’s law by mentioning the current flowing in the wire, without emphasizing on the source of the current in the wire. To apply Ampere’s circuital law, we need to have a constant current flowing in the wire, irrespective of its source.

Quite often, connecting wires carrying currents in opposite directions are twisted together in using electrical appliances. Explain how it avoids unwanted magnetic fields.

According to the diagram, we can see that the net magnetic field due to first turn is into the paper and due to second twisted turn is out of the plane of paper so these fields will cancel each other. Hence if the wire is twisted, then the resultant fields from consecutive twists are in opposite directions. So the cumulative effect over a long length of wire is roughly zero.

 

 

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