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Force On A Current-Carrying Wire In A Uniform Magnetic Field

Whenever a charge is moving, it’s suffering from a field of force. It creates a force, which is normal to its movement. This property of charge covers plenty of fields, as an example, this phenomenon is employed within the making of motors which successively are useful for creating mechanical forces. These forces are represented by the right-hand thumb rule and are given by the vector products. When a current-carrying wire is exposed to the flux it also experiences forces because the fees are moving inside the conductor. In this context, we will see the force between two parallel wires which are carrying current in a uniform magnetic field.

Force on a Moving Charge in a Magnetic Field :

The force experienced by the conductor is given by the Right-Hand Thumb Rule.  We can calculate the force on a current-carrying conductor and the force between two parallel wires by helping with this rule. This force  may be determined by taking an amount of magnetic attraction on the singular charges. Consider a single charge moving with a drift velocity vd. The following force  is acting on  this charge which is given by 

 F=qvBsin

Assume the force field B, to be uniform over the length “l” of the wire and nil wherever else. the entire magnetic attraction on the wire, all things considered, are going to be given by,F=qvBsin

Since, each charge is moving of equivalent speed, the all out force may be re-composed as,F=qvBsinN

Where N is the quantity of charges stricken by the flux. Suppose “n” is that the quantity of charge transporters per unit volume of the conveyors and “V” is that the volume of the realm of the wire where the field of force is acting.

N=nV

So,  FqvBsin()(nV)

F=qvBsin()nv

 

Likewise, since the wire is uniform the V = Al, where An is that the cross-sectional region and l is that the length of the wire under flux. Connecting this esteem the condition,F=qvBsin()nAl

We know that,

nqAv=i

 So the equation becomes F=ilBsin()

This force is given by in vector form F=i(LB)

 

Magnetic Field due to Current in Straight Wire :

As the charges are moving inside the conductor, a force field is produced by a current-carrying wire. This is further well checked  by a simple experiment of keeping a compass near any current-carrying wire. Different types and shapes of current-carrying conductors are available. The field produced is affected by the form of the conductor. To know and understand more, we have to go through the force between the parallel wires.

 

Magnetic Fields due to Moving Charges :

A field is always produced by moving charges. Charges are always run inside a current-carrying conductor and the magnetic fields are generated around them by such conductors. The sector that’s produced by these charges will be visualised. Right-hand thumb rule is used to provide the direction of the magnetic flux. In this right-hand thumb rule, the direction of the current is indicated by the thumb. And the direction of the field around the wire is indicated by the other four curled fingers.

 

Biot-Savart Law :

A magnetic field is always set-up around a current carrying wire. The intensity, at any point, in this magnetic field can be obtained with the help of “Biot-Savart’s Law.”

 Biot-Savart’s relation between electric current and magnetic field is given by

Magnetic Field due to a straight current-carrying wire : 

Let AB be an infinitely long conductor through which I amount of current flows. From the centre of the conductor, let there be a point P. Consider dl be the tiny current carrying element at point c at a distance r from point p. α be the angle between r and dl. l be the gap between centre of the coil and elementary length dl. From biot-savart law, field of force because of current carrying element dl at point P is

From the above three equations, 

B=μ[sin⁡(θ1)+sin⁡(θ2)]/4

Conclusion :

As we know, a current carrying wire creates a magnetic field around it which produces the magnetic force. We will predict these  by the famous formula given by Fleming. Hence the force between two parallel wires is best described in the above.

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Frequently asked questions

Get answers to the most common queries related to the IIT JEE Examination Preparation.

(Question1) What is Fleming’s Left-hand Rule?

Answer) Fleming’s Left-hand rule is employed to search out the direction of Force. If we put our left thumb, index...Read full

(Question2) Where B = 1.50 T, l = 5.00 cm, and I = 20.0 A. Current and Magnetic fields are inclined at 90 degrees with each other. Then calculate the force on the wire?

Answer) The force is  F=ilBsin(...Read full

(Question3) Where B = 3 T, l = 50.00 cm, and I = 10.0 A. The angle between the current and the magnetic field is 30°.What is the force on the wire ?

Answer) The force is F=ilBsin()...Read full