Question

The field strength midway between two parallel wires carrying equal currents in the same direction will be

Answer 1

The field strength midway between two parallel wires carrying equal currents in the same direction will be perpendicular to the plane of the two wires and non-zero. The Fleming's right hand rule is used in this situation and it states that to determine the direction of the magnetic force on a positive moving charge, ƒ, point the thumb of the right hand in the direction of motion, the first finger in the direction of the magnetic field, and the second finger perpendicular to the palm points in the direction of the current.

Answer 2

Answer:

ZERO

Explanation:

As we know that magnetic field due to a straight wire at some distance from the wire is given as

$B = \frac{\mu_0 i(\vec {dl} \times \hat r)}{4\pi r^2}$

so here since we have to find the magnetic field at the mid point of two parallel wire having equal current in them

so we can say

$B_1 = \frac{\mu_0 i(\vec {dl} \times \hat r)}{4\pi r^2}$

and since it is mid of of two wires so all the parameters will be same but the direction would be reversed as the mid point lies on the right side of one wire and on the left side of other wire.

$B_2 = \frac{\mu_0 i(\vec {dl} \times (-\hat r))}{4\pi r^2}$

now net magnetic field at mid point will be

$B = B_1 + B_2$

$B = 0$