Question

Two long, straight wires, each carrying a current of 5 A, are placed along the x- and y-axis respectively. The currents point along the positive directions of the axes. Find the magnetic fields at the points (a) (1 m, 1 m), (b) (−1 m, 1 m), (c) (−1 m, −1 m) and (d) (1 m, −1 m).

Answer 1

Answer:

this is the megnatic. field at the point

a 1m,-1m

Answer 2

Explanation:

To find the magnetic fields:

Given :

Current Magnitude field , I = 5 A

Separating of the point from the wire, d = 1 m

Thus the intensity of the magnetic field in the coil is defined by the current.

$B_{1}=B_{2}=\frac{\mu_{0} i}{2 \pi d}$

 (a) Due to the wires the magnetic fields at stage (1 m, 1 m) are the same in size but in the opposite path.

So the total average magnetic field is zero.

(b) The magnetic fields at point (−1 m, 1 m) are in upward direction because of the wires.

$B_{n e t}=B_{1}=B_{2}$

$=\frac{2 \times 10^{-7} \times 5}{1}+\frac{2 \times 10^{-7} \times 5}{1}$

$=10 \times 10^{-7}+10 \times 10^{-7}$

$=10^{-7}(10+10)$

$=20 \times 10^{-7}$

$=2 \times 10^{-6} T$   (By the z-axis)

(c) Due to the wires the magnetic fields at stage (-1 m, -1 m) are the same in size but in the opposite path.

So the total average magnetic field is zero.

(d) The magnetic fields are in the upward direction at point (1 m, −1 m), due to the wires.

$B_{n e t}=B_{1}=B_{2}$

= $=\frac{2 \times 10^{-7} \times 5}{1}+\frac{2 \times 10^{-7} \times 5}{1}$

=$=10 \times 10^{-7}+10 \times 10^{-7}$

= $=10^{-7}(10+10)$

= $=20 \times 10^{-7}$

= $2 \times 10^{-6} T$   (by the -ve z-axis)