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# Current Carrying Conductors Assignment Help

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Electromagnetism - Current Carrying Conductors

**Current Carrying Conductors**

Consider **C**_{1} D_{1} and **C**_{2} D_{2} two infinite long straight conductors carrying currents **I**_{1} ad **I**_{2} in the same directing they are held parallel to each other at a distance r apart in the plane of paper. The magnetic field is produce due to current through each conductor shown separately in and **(b)**. Since each conductor is in the magnetic field produced by the other therefore each conductor experiences a force.

Magnetic field induction at a point p one conductor **C**_{2} D_{2} due to current **I**_{1} passing through **C**_{1} D_{1} is given by **B**_{1} = μ_{0} / 4 π ^{2} I_{1} / r (1)

According to right hand rule the direction of magnetic field **B**_{1} is perpendicular to the plane of paper directed inwards.

As the current carrying conductor **C**_{2} D_{2} lies in the magnetic field B1 (produced by the current through **C**_{1}D_{1 }therefore, the unit length of **C**_{2} D_{2} will experience a force given by

**F**_{2} = B_{1} I_{2} X 1 = B_{1} I_{2 }

Putting the value of B1 we have

**F**_{2} = μ_{0} / 4 π ^{2} I _{1} I_{2} / r ** (2) **

According to Fleming's left hand rule force on conductor **C**_{2} D_{2} acts in the place of the paper perpendicular to **C**_{2} D_{2} directed towards **C**_{1} D_{1 }similarly it cane shown that the conductor **C**_{1} D_{1} also experiences a force given by eqn. (2) which acts into he lance of paper perpendicular to **C**_{1} D_{1} and directed towards **C**_{2} D_{2}hence **C**_{1} D_{1} and **C**_{2} D_{2} attract each other.

It means the two linear parallel conductors carrying currents in the same direction attract each other. The force experienced by the unit length of each conductor is given by (2) further it can be shown that if the current in conductor **C**_{1}D_{1} and **C**_{1} D_{2} are in opposite directions they repel each other with the same force as given by eq. (2)

**Definition of ampere**

Relation is used to define 1 ampere

Let **I**_{1} = I_{2} = IA; r = 1 m

**Then F = μ**_{0} / 4π ^{2}I_{1} I_{2} / r = 10^{-7} x 2 x 1 x 1 / 1

(∴ μ_{0} / 4 π = 10^{-7}) = 2 x 10^{ -7} Nm^{ - 1 }

Thus one ampere is that much current which when flowing through each of the two parallel uniform long linear conductor placed in free space at a distance of one meter from each other will attract or repel each other with a force of** 2 x 10**^{-7} N per meter of the length.

Note if two liner current carrying conductors of unequal length are held parallel to each other then the force on long conductor is due short conductor and long conductors.

if **I L** = length of short and long conductor respectively.

**I**_{1} I2 = current through short hang long conductor

R = separation between these two parallel conducts (i) then force on long conduct = force on short conductor = **μ**_{0} / 4 π^{2}I_{1} I_{2} / r.

**(ii)** The force on each conductor is same in magnitude and opposite in direction the force on two conductors obey Newton's third law of motion.

**(b) **When two currents approach a point or they go away from that point then they experience an attractive force.

**(c) **When one current out of the two approaches a point and another one goes away from that point they then experience a force of repulsion.

**(d) **When two charges of values **q**_{1} and **q**_{2} moving with velocities **v**_{1} and **v**_{2} on parallel straight paths distance r apart then the force acting between them is given by

**F = μ**_{0} / 4 π q_{1} q_{2} v_{1} v_{2} / r^{2 }

Proof: When a charge q moving with velocity v for time dt covers a distance dl it will behave as a current element I dl where

**Idl = q / dt dl = q dl / dt = qv**

If **I**_{1} I_{2} are the currents corresponding to motion of charges **q**_{1 }and **q**_{2} with velocity **v**_{1} and **v**_{2} then

**I**_{1} dl_{1} = q_{1} v and I_{2} dl _{2} = q_{2} v_{2}

The magnitude of magnetic force acting between these two parallel current elects distance **r** apart is given by

**Fm = μ**_{0} / 4π I_{1} I_{2} / r^{2} dl_{1} dl_{2} or Fm = μ_{0} q_{1 }q_{2} / 4π r^{2} v_{1} v_{2}.

**(e)** A fixed horizontal wire A carrying current** I**_{1} will support another horizontal wire B, carrying current **I**_{2}hanging in air at a distance r below the wire A if the direction of currents in the two wires is same here weight pr unit length of wire B is equal to the force of attraction acting per unit length of the wire B

**Mg μ**_{0} / 4π ^{2}I_{1} I_{2} / r

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**Current Carrying Conductors**

**C**and

_{1}D_{1}**C**

_{2}D_{2}two infinite long straight conductors carrying currents

**I**

_{1}ad

**I**

_{2}in the same directing they are held parallel to each other at a distance r apart in the plane of paper. The magnetic field is produce due to current through each conductor shown separately in and

**(b)**. Since each conductor is in the magnetic field produced by the other therefore each conductor experiences a force.

Magnetic field induction at a point p one conductor

**C**

_{2}D_{2}due to current

**I**

_{1}passing through

**C**

_{1}D_{1}is given by

**B**

_{1}= μ_{0}/ 4 π^{2}I_{1}/ r (1)According to right hand rule the direction of magnetic field

**B**

_{1}is perpendicular to the plane of paper directed inwards.

As the current carrying conductor

**C**

_{2}D_{2}lies in the magnetic field B1 (produced by the current through

**C**

_{1}D_{1 }therefore, the unit length of

**C**

_{2}D_{2}will experience a force given by

**F**

_{2}= B_{1}I_{2}X 1 = B_{1}I_{2 }

Putting the value of B1 we have

**F**

_{2}= μ_{0}/ 4 π^{2}I_{1}I_{2}/ r**(2)**

According to Fleming's left hand rule force on conductor

**C**

_{2}D_{2}acts in the place of the paper perpendicular to

**C**

_{2}D_{2}directed towards

**C**

_{1}D_{1 }similarly it cane shown that the conductor

**C**

_{1}D_{1}also experiences a force given by eqn. (2) which acts into he lance of paper perpendicular to

**C**

_{1}D_{1}and directed towards

**C**hence

_{2}D_{2}**C**

_{1}D_{1}and

**C**

_{2}D_{2}attract each other.

It means the two linear parallel conductors carrying currents in the same direction attract each other. The force experienced by the unit length of each conductor is given by (2) further it can be shown that if the current in conductor

**C**

_{1}D_{1}and

**C**

_{1}D_{2}are in opposite directions they repel each other with the same force as given by eq. (2)

**Definition of ampere**

Relation is used to define 1 ampere

Let

**I**

_{1}= I_{2}= IA; r = 1 m**Then F = μ**

(∴ μ

_{0}/ 4π^{2}I_{1}I_{2}/ r = 10^{-7}x 2 x 1 x 1 / 1(∴ μ

_{0}/ 4 π = 10^{-7}) = 2 x 10^{ -7}Nm^{ - 1 }

Thus one ampere is that much current which when flowing through each of the two parallel uniform long linear conductor placed in free space at a distance of one meter from each other will attract or repel each other with a force of

**2 x 10**per meter of the length.

^{-7}NNote if two liner current carrying conductors of unequal length are held parallel to each other then the force on long conductor is due short conductor and long conductors.

if

**I L**= length of short and long conductor respectively.

**I**= current through short hang long conductor

_{1}I2R = separation between these two parallel conducts (i) then force on long conduct = force on short conductor =

**μ**

_{0}/ 4 π^{2}I_{1}I_{2}/ r.**(ii)**The force on each conductor is same in magnitude and opposite in direction the force on two conductors obey Newton's third law of motion.

**(b)**When two currents approach a point or they go away from that point then they experience an attractive force.

**(c)**When one current out of the two approaches a point and another one goes away from that point they then experience a force of repulsion.

**(d)**When two charges of values

**q**

_{1}and

**q**

_{2}moving with velocities

**v**and

_{1}**v**

_{2}on parallel straight paths distance r apart then the force acting between them is given by

**F = μ**

_{0}/ 4 π q_{1}q_{2}v_{1}v_{2}/ r^{2 }

Proof: When a charge q moving with velocity v for time dt covers a distance dl it will behave as a current element I dl where

**Idl = q / dt dl = q dl / dt = qv**

If

**I**are the currents corresponding to motion of charges

_{1}I_{2}**q**

_{1 }and

**q**

_{2}with velocity

**v**and

_{1}**v**

_{2}then

**I**

_{1}dl_{1}= q_{1}v and I_{2}dl_{2}= q_{2}v_{2}

The magnitude of magnetic force acting between these two parallel current elects distance

**r**apart is given by

**Fm = μ**

_{0}/ 4π I_{1}I_{2}/ r^{2}dl_{1}dl_{2}or Fm = μ_{0}q_{1 }q_{2}/ 4π r^{2}v_{1}v_{2}.

**(e)**A fixed horizontal wire A carrying current

**I**

_{1}will support another horizontal wire B, carrying current

**I**hanging in air at a distance r below the wire A if the direction of currents in the two wires is same here weight pr unit length of wire B is equal to the force of attraction acting per unit length of the wire B

_{2}**Mg μ**

_{0}/ 4π^{2}I_{1}I_{2}/ rExpertsMind.com - Current Carrying Conductors Assignment Help, Current Carrying Conductors Homework Help, Current Carrying Conductors Assignment Tutors, Current Carrying Conductors Solutions, Current Carrying Conductors Answers, Electromagnetism Assignment Tutors