Question

Three long parallel wires each carry 2.0-A currents in the same direction. The wires are oriented vertically, and they pass through three of the corners of a horizontal square of side 4.0 cm. What is the magnitude of the magnetic field at the fourth (unoccupied) corner of the square due to these wires? (μ0 = 4???? × 10-7 T · m/A)

Answer 1

Answer:

$21.2\times 10^{-6}$ T

Explanation:

$i$  = magnitude of current in each wire = 2.0 A

$a$  = length of the side of the square = 4 cm = 0.04 m

$r$  = length of the diagonal of the square = $\sqrt{2}$ a = $\sqrt{2}$ (0.04) = 0.057 m

$B$ = magnitude of magnetic field by wires at A and C

$B = \left ( \frac{\mu _{o}}{4\pi } \right )\left ( \frac{2i}{a} \right )$

$B = (10^{-7}) \left ( \frac{2(2)}{0.04} \right )$

$B = 10\times 10^{-6}$ T

$B'$ = magnitude of magnetic field by wire at B

$B' = \left ( \frac{\mu _{o}}{4\pi } \right )\left ( \frac{2i}{r} \right )$

$B' = (10^{-7}) \left ( \frac{2(2)}{0.057} \right )$

$B' = 7.02\times 10^{-6}$ T

Net magnitude of the magnetic field at D is given as

$B_{net} = \sqrt{B^{2}+B^{2}} + B'$

$B_{net} = \sqrt{2} B + B'$

$B_{net} = \sqrt{2} (10\times 10^{-6}) + (7.02\times 10^{-6})$

$B_{net} = 21.2\times 10^{-6}$ T