Answered: The figure below shows two vertical,…

Glencoe Physics: Principles and Problems, Student Edition

1st Edition

ISBN:9780078807213

Author:Paul W. Zitzewitz

Publisher:Paul W. Zitzewitz

Chapter24: Magnetic Fields

Section: Chapter Questions

Problem 79A

See similar textbooks

Related questions

Question

ASK YOUR TEACHER PRACTICE ANOTHER The figure below shows two vertical, parallel wires separated by a distance d = 11.5 cm. The left wire carries a current of I = 5.20 A upward, while the right wire carries a current of the same magnitude directed downward. Point P1 is a distance d to the right of the right wire, and point P2 is a distance 2d to the left of the left wire. Two vertical, parallel wires are separated by a distance d. To the left of the left wire is an arrow labeled I pointing up. To the right of the right wire is an arrow labeled I pointing down. A point P2 is a distance 2d to the left of the left wire, and a point P1 is a distance d to the right of the right wire. (a) What are the magnitude (in T) and direction of the net magnetic field at a point midway between the wires? magnitude T direction (b) What are the magnitude (in T) and direction of the net magnetic field at point P1, to the right of the right wire? magnitude T direction (c) What are the magnitude (in T) and direction of the net magnetic field at point P2, to the left of the left wire? magnitude T direction

4:57
A webassign.net
JIstancE d = 20.0 cm apa
|3D
e the +x-direction is
I2
nat are the magnituc
agnetic field due to t
res? Measure the ang
e midpoint at the orie
Id due to the Earth.)

Expert Solution

Step 1

Given: The separation between the two vertical wires is 11.5 cm.

The current of the left wire is 5.20 A in the upward direction.

The current of the right wire is 5.20 A in the downward direction.

To determine: The direction and magnitude of net magnetic field at a midpoint between the two wires.

The magnetic field due to the wire is

$B = \frac{μ_{0} I}{2 πr}$

where B is the magnetic field, I is the current and r is the distance.

Right in the middle, the magnetic field will add up and the current is same, so

$B_{m i d d l e} = 2 \times \frac{μ_{0} I}{2 πr}$

Substitute 5.20 A for I, $4 π \times 10^{- 7} N / A^{2}$ for $μ_{0}$ and 0.115 m for r in the above equation,

$\begin{array}{rcl} B_{middle} & = & 2 \times \frac{4 π \times 10^{- 7} N / A^{2} \times 5.20 A}{2 π \times 0.115 m} \\ = & 1.808 \times 10^{- 5} T \end{array}$

The direction of net magnetic field is into the page.

(b) The magnitude and direction of the net magnetic field at point $P_{1}$ ,

The magnetic field at this point will subtract. According to the Fleming's right hand rule, place the thumb in the direction of current then curl the fingers, the direction of fingers will give the direction of magnetic field.

The magnitude of the magnetic field at point $P_{1}$ is

$B_{P} = \frac{μ_{0} I}{2 π d} - \frac{μ_{0} I}{2 π (2 d)}$

Substitute 5.20 A for I, $4 π \times 10^{- 7} N / A^{2}$ for $μ_{0}$ and 0.115 m for d in the above equation,

$\begin{array}{rcl} B_{P} & = & \frac{4 π \times 10^{- 7} N / A^{2} \times 5.20 A}{2 π \times 0.115 m} - \frac{4 π \times 10^{- 7} N / A^{2} \times 5.20 A}{2 π \times (2 \times 0.115 m)} \\ = & \frac{4 π \times 10^{- 7} N / A^{2} \times 5.20 A}{2 π \times 0.115 m} (1 - \frac{1}{2}) \\ = & 45.2 \times 10^{- 7} T \end{array}$

The direction of the net magnetic field is out of the page.

Trending now

This is a popular solution!

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.