Let P and Q be the two conductors, d metres apart, carrying current I_{1} and I_{2 }amperes as shown in Fig.16.

The flux density Q due to P,

If l is the length of conductor Q lying in this flux density, then force on this conductor,

F = BI_{2}l_{2} newtons

(If the currents I_{1} and I_{2} are in the same direction, then the force between the conductors P and Q is one attraction ; whereas if the current are in opposite direction, the conductors repel each other). Since Eqn. (9) is known as Ampere’s Law and is used to define the ampere in S.I. units.

µ_{0} = 4π × 10^{-7} H/m

If, l = d= 1 m,

and, I_{1} = I_{2} = 1A

then F = 2 × 10^{-7} N

Hence, one ampere current is defined as follows:

*“An ampere is that current which when flowing in each of the two infinitely long parallel conductors situated in vacuum and separated 1 meter between centres, produces on each conductor a force of 2×10 ^{-7} N per meter length.”*

**Example 4. ***Two long parallel conductors carrying current, in the same direction, of 2500 A and 5000 A are placed with their axis 100 mm apart. Calculate the force between them in N/m.*

**Solution, **Current, I_{1} = 2500 A

Currents I_{2} = 5000 A

Distance between the conductors d = 100 mm = 0.1 m.

Force between the conductors per metre, F:

**= ****25 ****N/m. ****(Ans.) **

**Example 5**. *Two long straight parallel wires, standing in air *2 *m apart, carry currents I _{1} and I_{2} in the same direction. The magnetic intensity at a point midway between the wires is *7.95

*AT*/

*m. If the force on each wire per unit length is*2.4 x

*10-*N,

^{4}*evaluate I*

_{1}and I_{2}:**(Madras University) **

**Solution**. Distance between the conductors, *d *= 2 m

Magnetic intensity midway between the wires = 7.95 AT/m

Force on each wire per unit length = 2.4 x 10^{-4} N

**I _{1} , I_{2} :**

We know that when the two currents flow in the same direction, net field strength midway between the two conductors is the difference of the two field strengths.

or I_{1} -1_{2} = 2π × 7.95 = 50

Also using the relation,

Substituting the value of *I _{1} *from

*(i)*in

*(ii),*we get

(50 + *I _{2}) *x

*I*= 2400

_{2}or I_{2}^{2} + 50I_{2} – 2400 = 0

**Hence I _{1} = 80 A and I_{2} = 30 A. (Ans.)**

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