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

The correct option is a.

The magnitude of the magnetic field at the fourth corner of the square due to the wires is 2.1 µT.

To calculate the magnitude of the magnetic field at the fourth corner of the square, we can use the Biot-Savart law. Since the wires are long, the magnetic field at the unoccupied corner will follow the formula:

B = (μ0 x I) / (2π x r)

where B is the magnetic field, μ0 is the permeability of free space, I is the current, and r is the distance from the wire to the unoccupied corner. In this case, the current is 2.0 A, and the distance is the diagonal of the square with side 4.0 cm. Using Pythagoras' theorem, we find that the distance is approximately 5.66 cm.

Plugging these values into the formula, we get:

B = (4π x 10-7 T • m/A x 2.0 A) / (2π x 5.66 x 10-2 m) = 2.1 µT

Therefore, the magnitude of the magnetic field at the fourth corner of the square is 2.1 µT, which is Option A.

The complete question is here:

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 = 4π × 10-7 T • m/A)

A)2.1 µT

B)21 µT

C)1.2 µT

D) 0 T

E)12 µT

Answer 2

`21.2* 10^(-6)` T

Step-by-step 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 =
`√(2)` a =
`√(2)` (0.04) = 0.057 m

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

`B = \left ( (\mu _(o))/(4\pi ) \right )\left ( (2i)/(a) \right )`

`B = (10^(-7)) \left ( (2(2))/(0.04) \right )`

`B = 10* 10^(-6)` T

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

`B' = \left ( (\mu _(o))/(4\pi ) \right )\left ( (2i)/(r) \right )`

`B' = (10^(-7)) \left ( (2(2))/(0.057) \right )`

`B' = 7.02* 10^(-6)` T

Net magnitude of the magnetic field at D is given as

`B_(net) = \sqrt{B^(2)+B^(2)} + B'`

`B_(net) = √(2) B + B'`

`B_(net) = √(2) (10* 10^(-6)) + (7.02* 10^(-6))`

`B_(net) = 21.2* 10^(-6)` T