हिंदी

Two Infinitely Long Straight Parallel Wires, '1' and '2', Carrying Steady Currents I1 and I2 in the Same Direction Are Separated by a Distance D. Obtain the Expression for the Magnetic Field `B`Due to the Wire '1' Acting on Wire '2'. Hence Find Out, with the Help of a Suitable Diagram, - Physics

Advertisements
Advertisements

प्रश्न

Two infinitely long straight parallel wires, '1' and '2', carrying steady currents I1 and I2 in the same direction are separated by a distance d. Obtain the expression for the magnetic field `vecB`due to the wire '1' acting on wire '2'. Hence find out, with the help of a suitable diagram, the magnitude and direction of this force per unit length on wire '2' due to wire '1'. How does the nature of this force changes if the currents are in opposite direction? Use this expression to define the S.I. unit of current.

Advertisements

उत्तर

Consider a straight conductor XY lying in the plane of paper. Consider a point P at a perpendicular distance a from straight conductor.

Magnetic field induction (B) at a point P due to current I passing through conductor XY is given by

`B=(μ0I)/(4πa)[sinϕ_1+sinϕ_2]`

where ϕ1 and ϕ2  are the angles made by point X and Y, respectively

At the centre of the infinite long wire, ϕ1=ϕ2=90°

`B=(μ_0I)/(4πa)[sin90^@+sin^@]`

`⇒B=μ_0/(4π) (2I)/a   .....(1)`

Let 1 and 2 be two long infinite straight conductors. Let I1 and I2 be the current flowing through the conductor 1 and 2 and they are d distance apart from each other as shown in the figure.

The magnetic field induction (B) at a point P on conductor 2 due to current I1 passing through conductor 1 is given by

`B=(μ_02I_1)/(4πd)  ` [From (1), where a=d]

According to right hand rule, the direction of this magnetic field is perpendicular to the plane of the paper inward.
Since the conductor 2 lies in this magnetic field of conductor 1, force experienced (F2) by unit length of conductor 2 will be

F2=B1I2×1=B1I2

`∴ F2=μ_0/(4π) (2I_1I_2)/d`

Conductor 1 also experiences the same amount of force, directed towards the conductor 2. Hence, conductor 1 and conductor 2 attract each other. Thus, two linear parallel conductors carrying currents in the same direction attract and repel each other, when the current flows in the opposite direction.

Let I_1=I_1=1A; r=1 m
Then,

`F_1=F_2=F=10^(−7) (2xx1xx1)/1`

`⇒F=2×10^(−7)  N/m`

Thus, one ampere is that value of constant current which when flowing through each of the two parallel uniform long linear conductors placed in free space at a distance of 1 m from each other will attract or repel each other with a force of 2 × 10−7 N per metre of their length.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
2014-2015 (March) Patna Set 2

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

A charged particle goes undeflected in a region containing an electric and a magnetic field. It is possible that
(a) `vecE" || "vecB , vecv" || " vec E `
(b) `vecE  "is not parallel"  vecB`
(c) `vecv " || " vecB  but  vecv  "is not parallel"`
(d) `vecE" || " vecB  but   vecv "is not parallel"`


An electron is moving along the positive x-axis. You want to apply a magnetic field for a short time so that the electron may reverse its direction and move parallel to the negative x-axis. This can be done by applying the magnetic field along
(a) y-axis
(b) z-axis
(c) y-axis only
(d) z-axis only


A long, straight wire carries a current along the z-axis, One can find two points in the xy 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.



The magnetic field existing in a region is given by  `vecB = B_0(1 + x/1)veck` . A square loop of edge l and carrying a current i, is placed with its edges parallel to the xy axes. Find the magnitude of the net magnetic force experienced by the loop.


Four long, straight wires, each carrying a current of 5.0 A, are placed in a plane as shown in figure. The points of intersection form a square of side 5.0 cm.
(a) Find the magnetic field at the centre P of the square.
(b) Q1, Q2, Q3, and Q4, are points situated on the diagonals of the square and at a distance from P that is equal to the diagonal of the square. Find the magnetic fields at these points. 


A long, straight wire carries a current i. Let B1 be the magnetic field at a point P at a distance d from the wire. Consider a section of length l of this wire such that the point P lies on a perpendicular bisector of the section B2 be the magnetic field at this point due to this second only. Find the value of d/l so that B2 differs from B1 by 1%.    


Define Ampere in terms of force between two current carrying conductors.


Two free parallel wires carrying currents in the opposite directions ______.

The magnetic moment of a circular coil carrying current is ______.

Two long straight parallel conductors carrying currents I1 and I2 are separated by a distance d. If the currents are flowing in the same direction, show how the magnetic field produced by one exerts an attractive force on the other. Obtain the expression for this force and hence define 1 ampere.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×