Question

two long parallel conductors carry 100 a. if the conductors are separated by 20 mm, the force per metre of length of each conductor will be

Answer 1

Answer:

0.1N

Explanation: Current I1 and I2=100A

Distance =20mm

Force=?

Torque developed force, fT=MO*I1*I2/2*(pi)*d

=4*(pi)*10^-7*100^2/2*(pi)20*10^-3

=0.1N

Answer 2

Concept:

The force per unit length between two long parallel conductors carrying current $I_{1}$ and $I_{2}$ which is separated by distance $r$. This force is attractive if current flows in same direction and if current flows in reverse direction then this force is repulsive.

Given:

The current $I_{1}$ and $I_{2}$ are $100 a$, separated by distance $r=20mm=20*10^{-3}m$.

To Find:

We have to find the force per unit length of each conductor.

Solution:

The formula for force per unit length is given by $F=\frac{u_{0} I_{1}I_{2} }{2\pi r}$

Putting the value from given data,

$F=\frac{4\pi *10^{-7}*100*100 }{2\pi *20*10^{-3} }$

$F=0.1 N$

Hence the force per unit length of each conductor is $0.1 N$

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