/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 Two long parallel wires \(P\) an... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 19

Two long parallel wires \(P\) and \(Q\) are held perpendicular to the plane of the paper at a separation of \(5 \mathrm{~m}\). If \(P\) and \(Q\) carry currents of \(2.5 \mathrm{~A}\) and \(5 \mathrm{~A}\), respectively, in the same direction, then the magnetic field at a point midway between \(P\) and \(Q\) is (A) \(\frac{\mu_{0}}{\pi}\) (B) \(\frac{\sqrt{3} \mu_{0}}{\pi}\) (C) \(\frac{\mu_{0}}{2 \pi}\) (d) \(\frac{3 \mu_{0}}{2 \pi}\)

Short Answer

Expert verified
The short answer based on the given step-by-step solution is: The magnetic field at a point midway between P and Q is \(\frac{3\mu_{0}}{2 \pi}\).

Step by step solution

01

Biot-Savart Law

The Biot-Savart Law is given by $$ B=\frac{\mu_{0}}{4 \pi} \frac{I \cdot dl \times \hat {r}}{r^2} $$ where \(B\) is the magnetic field, \(\mu_0\) is the permeability of free space, \(I\) is the current, \(dl\) is the differential length of the wire, \(\hat{r}\) is a unit vector from the wire to the required point, and \(r\) is the distance between the wire and the required point. **Step 2: Determine the magnetic field due to wire P.**
02

Magnetic field due to wire P

Let the point midway between the wires be M. Since we are given that the wires are perpendicular to the plane of the paper, we can compute the magnetic field MBA due to wire P at this point using Biot-Savart law. The distance r between wire P and point M is half of their separation, i.e., \(\frac{5\mathrm{~m}}{2}=2.5\mathrm{~m}\). Thus, the magnetic field due to wire P at point M is: $$ B_{P}=\frac{\mu_{0} \cdot I_{P}}{2 \cdot 2.5\pi}=\frac{\mu_{0}}{2\pi} $$ **Step 3: Determine the magnetic field due to wire Q.**
03

Magnetic field due to wire Q

Similarly, we can compute the magnetic field MBQ due to wire Q at point M using Biot-Savart law. The distance r between wire Q and point M is also 2.5 m. Thus, the magnetic field due to wire Q at point M is: $$ B_{Q}=\frac{\mu_{0} \cdot I_{Q}}{2 \cdot 2.5\pi}=\frac{\mu_{0}}{\pi} $$ **Step 4: Determine the net magnetic field at point M.**
04

Net Magnetic field at point M

Since both the magnetic fields are in the same direction, we will add them linearly to obtain the net magnetic field at point M: $$ B_{M}=B_{P} + B_{Q}=\frac{\mu_{0}}{2 \pi}+\frac{\mu_{0}}{\pi} $$ $$ B_{M}=\frac{3\mu_{0}}{2 \pi} $$ Hence, the magnetic field at a point midway between P and Q is \(\frac{3\mu_{0}}{2 \pi}\). Therefore, the correct option is (d).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Three identical bar magnets, each of magnetic moment \(M\), are placed in the form of an equilateral triangle with north pole of one touching the south pole of the other as shown. The net magnetic moment of the system is (A) Zero (B) \(3 M\) (C) \(\frac{3 M}{2}\) (D) \(M \sqrt{3}\)

Earth's magnetic induction at a certain point is \(7 \times 10^{-5}\) \(\mathrm{Wb} / \mathrm{m}^{2}\). This field is to be annulled by the magnetic induction at the centre of a circular conducing loop \(5.0 \mathrm{~cm}\) in radius. The required current is (A) \(0.056 \mathrm{~A}\) (B) \(6.5 \mathrm{~A}\) (C) \(5.6 \mathrm{~A}\) (D) \(12.8 \mathrm{~A}\)

A current \(i\) ampere flows in a circular arc of wire whose radius is \(R\), which subtend an angle \(3 \pi / 2\) radian at its centre. The magnetic induction \(B\) at the centre is (A) \(\frac{\mu_{0} i}{R}\) (B) \(\frac{\mu_{0} i}{2 R}\) (C) \(\frac{2 \mu_{0} i}{R}\) (D) \(\frac{3 \mu_{0} i}{8 R}\)

An electron of mass \(m\) is accelerated through a potential difference of \(V\) and then it enters a magnetic field of induction \(B\) normal to the lines. Then the radius of the circular path is (A) \(\sqrt{\frac{2 e V}{m}}\) (B) \(\sqrt{\frac{2 V m}{e B^{2}}}\) (C) \(\sqrt{\frac{2 V m}{e B}}\) (D) \(\sqrt{\frac{2 V m}{e^{2} B}}\)

A long straight wire carrying a current of \(30 \mathrm{~A}\) is placed in an external uniform magnetic field of magnitude \(4 \times\) \(10^{-4} \mathrm{~T}\). The magnetic field is acting parallel to the direction of current. The magnitude of the resultant magnetic field in tesla at a point \(2.0 \mathrm{~cm}\) away from the wire is (A) \(10^{-4}\) (B) \(3 \times 10^{-4}\) (C) \(5 \times 10^{-4}\) (D) \(6 \times 10^{-4}\)
See all solutions

Recommended explanations on Physics Textbooks

Dynamics

Read Explanation

Classical Mechanics

Read Explanation

Rotational Dynamics

Read Explanation

Torque and Rotational Motion

Read Explanation

Circular Motion and Gravitation

Read Explanation

Electric Charge Field and Potential

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.