Question

two parallel wire carry current of 10 a each in the same direction if the separation of between them is 2m then the force unit length of each wire is​

Answer 1

Answer:

Force on a current carrying conductor is F=∫i( dl × B )

Magnetic field due to one wire at the location of the second wire ( other one ) is B 1 = 2πd/μ0i1

Force on the 2nd wire due to B1 is F21 =

i2l2B1 =i2l2μ0i1 / 2πd = μ0i1i2l2 / 2πd

Force per unit length for wire l2 is

F per unit length = F21 / I2

B1 is in − k^direction.

l2 is in j^direction

Hence, F21 will be in j^×( k^)= i^ direction.

∴ it is an repulsive force.

F21 = 2×10-7× 10×2×2 / 10×10-2=8×10 −5 N

Here is your answer