Question

The two wires shown in the figure carry currents of 5.00 A in opposite directions and are separated by 10.0 cm. Find the direction and magnitude of the net magnetic field (a) at a point midway between the wires, (b) at point P1 (10.0 cm to the right of the wire on the right), and (c) at point P2 (20.0 cm to the left of the wire on the left). Ans: a=4uT into the page, b=5uT out of the page, c=1.67uT out of the page

Name: The two wires shown in the figure carry currents of 5.00 A in opposite directions and are separated by 10.0 cm. Find the direction and magnitude of the net magnetic field (a) at a point midway between the wires, (b) at point P1 (10.0 cm to the right of the wire on the right), and (c) at point P2 (20.0 cm to the left of the wire on the left). Ans: a=4uT into the page, b=5uT out of the page, c=1.67uT out of the page
Uploaded: 2022-04-07T04:45:41-08:00
Duration: 4 min 57 s
Channel: Darshan Maheshwari

          The two wires shown in the figure carry currents of 5.00 A in opposite directions and are separated by 10.0 cm. Find the direction and magnitude of the net magnetic field (a) at a point midway between the wires, (b) at point P1 (10.0 cm to the right of the wire on the right), and (c) at point P2 (20.0 cm to the left of the wire on the left). Ans: a=4uT into the page, b=5uT out of the page, c=1.67uT out of the page

The two wires shown in the figure carry currents of 5.00 A in opposite directions and are separated by 10.0 cm. Find the direction and magnitude of the net magnetic field (a) at a point midway between the wires, (b) at point P1 (10.0 cm to the right of the wire on the right), and (c) at point P2 (20.0 cm to the left of the wire on the left). Ans: a=4uT into the page, b=5uT out of the page, c=1.67uT out of the page

Added by Margarita J.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Darshan Maheshwari

04/07/2022

Step 1

Given: Current, I = 5 A Distance between the wires, d = 10 cm = 0.1 m Using the formula for magnetic field due to a current-carrying wire: \[ B = \frac{\mu_0 I}{2\pi r} \] For the wire on the left: \[ B_1 = \frac{\mu_0 I}{2\pi \times 0.05} = \frac{4\pi \times Show more…

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