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Force on a Current Carrying Conductor in a Magnetic Field

We know moving charge experiences a force due to magnetic field. Magnetic field creates force into conductor passing current through it. The charges of the conductor experience of force for magnetic field. Now we determine the force on a current carrying conductor in a magnetic field.

Force on a Current Carrying Wire in a Magnetic Field

Figure: Force on a Current Carrying Conductor in a Magnetic Field

Consider a network which is filled with magnetic field, B. The direction of magnetic field is perpendicular with the screen which is shown in cross sign in figure. A conductor carries I current from left to right side. Current flow left to right side that means flow of electron direction is right to left side.

Length of the conductor = l

Cross section area of the conductor = A

Number of electron per cube in the conductor = n

Charge per electron = q

Escape velocity of electron = v

Magnitude of magnetic field = B

Flow of current via conductor = i

As the conductor is placed into magnetic field all the electron of the conductor will experience individual force. Magnetic field direction and flow of current direction is perpendicular which is 90 degree angle between them.

Force on a single electron in the conductor due to magnetic field,

Fm = qvB sin90° = qvB

Fm means magnetizing force on a single electron. If the total number of electron is N for whole length of the conductor. Total force on the conductor for magnetic field,

F = N Fm

Here,

N = n × volume of conductor

= n Al

Total force, F = n Al Fm

= nAl qvB

We know for escape velocity or drift velocity of electron, i = nAvq

Hence, F = i l B

If current carrying conductor not placed perpendicular with magnetic field it is placed with another angle Ѳ then force on a single electron,

Fm = qvBsinѲ

Total force on whole conductor,

F = i l B sinѲ

Vector form of this equation is which indicates the direction of force and magnitude also

Vector form of length, l indicates the length of the conductor. The direction of l is considered the direction of flow of current.

If the conductor has N turns coil then the force created by magnetic field will be

Total force can be enhanced by increasing the magnitude of magnetic field or flow of current or increasing the number of turns or enhancing length of the conductor.

Direction of force created on current carrying conductor due to magnetic field

Direction of force on current carrying conductor created by magnetic field described by Flemming Left hand rule.

Flemming Left hand rule:

Case 1: If direction of current flow is perpendicular with magnetic field direction then created force direction can be explained by Flemming left hand rule. Flemming explained magnetic field direction, current flow direction and created force direction with three fingers of left hand. These left hand fingers are index finger, middle finger and thumb.

three fingers of Flemming left hand

To understand the direction take your left hand and point it as you are firing something with gun like below picture.

Now release the middle finger as below image.

Flemming showed his hand exactly same way above figure. To find the direction of creating force place index finger at the magnetic field direction and middle finger for the direction of current. Then thumb shows the creating force direction.

Figure: Flemming Left Hand Rule

Case 2: If the direction of current flow and magnetic field are parallel to each other that means there is no angle between them such Ѳ = 0° or Ѳ = 180°.

So that the current carrying conductor will not experience any force.

F = ilB sin0° = 0

Example: A 50cm 2 A current carrying conductor is placed in a 10-2 Tesla magnetic field and angle between conductor and magnetic field 60 degree. Find the force the current carrying conductor experienced.

Answer:

Here,

Magnetic field, B = 10-2 Tesla

Current flow through conductor, i = 2A

Length of the conductor, l = 50cm = 0.50m

Angle betn magnetic field and conductor, Ѳ = 60 degree

Forced created, F = ?

We know,

F = i l B sinѲ

= 2 × 0.50 × 10-2 ×sin60°

= 8.66 × 10-3 N

The force created due to flow of current in the magnetic

 

 

 

 

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