/*! 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} Q13P In Fig. 29-44 point P1 is at di... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In Fig. 29-44 point P1is at distance R=13.1cmon the perpendicular bisector of a straight wire of length L=18.0cm. carrying current. (Note that the wire is notlong.) What is the magnitude of the magnetic field at P1due to i?

Short Answer

Expert verified

The magnitude of the magnetic field at P1due to i is5.03×10-8T

Step by step solution

01

Given

  1. Current i=58.2mA
  2. Distance of point P on the perpendicular bisector of wireR=13.1cm=0.131m
  3. Length of wireL=18.0cm=0.18m
02

Determine the formulas:

Use the formula of Biot-Savarts law to find the magnitude of the field at P1due to i

Formula:

dB=μ0idl4πsinθr2

03

Calculate the magnitude of the magnetic field at P1 due to i

According to Bio-Savarts law

dB=μ0idl4πsinθr2

If r makes an angle θwithl then

r=l2+R2

And

sinθ=RR=Rl2+R2l=L2=0.18m2=0.09m

Integrating an equation of Bio-Savarts law

B=∫00.09dB=∫00.09μ0idl4πsinθr2B=μ0i4π∫00.09sinθdlr2B=μ0i4π∫00.09Rl2+R2dll2+R22B=μ0iR4π∫00.09dll2+R23/2

Solve further as:

B=μ0iR4π1R2ll2+R21/200.09B=μ0i2πRll2+R21/2B=4π×10-7TmA0.0582A2π0.131m0.09m0.09m2+0.131m21/2B=5.03×10-8T

Therefore, the magnitude of the magnetic field at P1due to i is5.03×10-8T

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 91Ó°ÊÓ!

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

A long vertical wire carries an unknown current. Coaxial with the wire is a long, thin, cylindrical conducting surface that carries a current of30 mAupward. The cylindrical surface has a radius of3.0 mm. If the magnitude of the magnetic field at a point5.0 mmfrom the wire is1.0μ°Õ, (a) What are the size and(b) What is the direction of the current in the wire?

In Fig. 29-44, point P2is at perpendicular distanceR=25.1cm from one end of a straight wire of length L=13.6cmcarrying current i=0.693A.(Note that the wire is notlong.) What is the magnitude of the magnetic field at P2?

A long, hollow, cylindrical conductor (with inner radius 2.0mm and outer radius 4.0mm) carries a current of 24A distributed uniformly across its cross section. A long thin wire that is coaxial with the cylinder carries a current of 24A in the opposite direction. What is the magnitude of the magnetic field (a) 1.0mm,(b) 3.0mm, and (c) 5.0mm from the central axis of the wire and cylinder?

Figure 29-30 shows four circular Amperian loops (a, b, c, d) concentric with a wire whose current is directed out of the page. The current is uniform across the wire’s circular cross section (the shaded region). Rank the loops according to the magnitude of ∮B→.ds→around each, greatest first.

In Fig. 29-40, two semicircular arcs have radii R2=7.80cmand R1=3.15cmcarry current i=0.281A and have the same center of curvature C. What are the (a) magnitude and (b) direction (into or out of the page) of the net magnetic field at C?

See all solutions

Recommended explanations on Physics Textbooks

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.