English

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,

Advertisements
Advertisements

Question

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

Solution

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
  Is there an error in this question or solution?
2014-2015 (March) Patna Set 2

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Two long straight parallel conductors 'a' and 'b', carrying steady currents Ia and Ib are separated by a distance d. Write the magnitude and direction of the magnetic field produced by the conductor 'a' at the points along the conductor 'b'. If the currents are flowing in the same direction, what is the nature and magnitude of the force between the two conductors?


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


A straight wire of length l can slide on two parallel plastic rails kept in a horizontal plane with a separation d. The coefficient of friction between the wire and the rails is µ. If the wire carries a current i, what minimum magnetic field should exist in the space in order to slide the wire on the rails?


Figure shows a metallic wire of resistance 0.20 Ω sliding on a horizontal, U-shaped metallic rail. The separation between the parallel arms is 20 cm. An electric current of 2.0 µA passes through the wire when it is slid at a rate of 20 cm s−1. If the horizontal component of the earth's magnetic field is 3.0 × 10−5 T, calculate the dip at the place.


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 which is 2.0 cm away from each of these wires.


Two parallel wires separated by a distance of 10 cm carry currents of 10 A and 40 A along the same direction. Where should a third current by placed so that it experiences no magnetic force?


Answer the following question.
Two infinitely long straight wire A1 and A2 carrying currents I and 2I flowing in the same direction are kept' distance apart. Where should a third straight wire A3 carrying current 1.5 I be placed between A1 and A2 so that it experiences no net force due to A1 and A2? Does the net force acting on A3 depend on the current flowing through it?


According to Ampere's circuital law, ______.


Five long wires A, B, C, D and E, each carrying current I are arranged to form edges of a pentagonal prism as shown in figure. Each carries current out of the plane of paper.

  1. What will be magnetic induction at a point on the axis O? AxisE is at a distance R from each wire.
  2. What will be the field if current in one of the wires (say A) is switched off?
  3. What if current in one of the wire (say) A is reversed?

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×