Question

Two long straight wires carry currents and l2 as shown. Indicate on the sketch the points at which the B-field vanishes. Indicate the direction and calculate the magnitude of at points a and b 个

Answer 1

We know magnetic field due to a long current carrying conductor at a disance is

$B= \frac{\mu I}{2�\pi r}$ where r is distance between point and wire, I is current and $\mu$ is free space permitivity = $4 \pi � 10^-7$ h/m and direction would be in the direction of roation of right handed scrue where direction of movement of scrue represents the current flow direction.

So at point a, the field due to I1 is = B1a=(4×π×10^-7×I1)/(2×π× (L/2) = (2×10^-7×I1)/(L/2) which is into the plane of the paper

at point a, the field due to I2 is = B2a=(4×π×10^-7×I2)/(2×π× (L/2) = (2×10^-7×I2)/(L/2) which is out the plane of the paper

Therefore net field at a = B1a~B2a the direction would be in the direction of which field has greater value, please put the value of I1 and I2. HERE I1 and I2 is not clear in question.

Now field at b, here for I1 length = 3×(L/2), the field due to I2 is = B1b=(4×π×10^-7×I2)/(2×π× (3L/2) = (2×10^-7×I1)/(3L/2) which is into the plane of the paper.

the field due to I2 at b is = B2b=(4×π×10^-7×I1)/(2×π× (L/2) = (2×10^-7×I2)/(L/2) which is also into the plane of the paper. So net field at b would be additive i.e.

B1b+B2b = (2×10^-7){(2×I1/3L)+(2×I2/L)} T which will be in the direction of into the plane of the paper. Please put the value of I1 and I2 as given accordingly.