A straight conductor carrying current i=5.0Asplits into identical semicircular arcs as shown in Figure. What is the magnetic field at the center C of the resulting circular loop?

Short Answer

Expert verified

The magnetic field at the center isB=0.

Step by step solution

01

Given Information

Angle made by arc=π

02

Determining the formula for the magnetic field

Formula:

Magnetic field at center of circular arc is given by:

BC=μ04π×iϕR

Here, ϕis the angle of arc and R is the radius of the arc.

03

Calculating the magnetic field at the center C of the resulting circular loop 

Magnetic field at center due to arc is given by:

B=μ04π×iϕR

Fields due to both arcs are opposite to each other.

Angle due to semicircular arcs =π,

Net field due to both semicircular arcs is

B=μ04π×iπR+μ04π×i-πR

B=0

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 Vaia!

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

In Figure, four long straight wires are perpendicular to the page, and their cross sections form a square of edge length a=13.5cm. Each wire carries7.50A, and the currents are out of the page in wires 1 and 4 and into the page in wires 2 and 3. In unit vector notation, what is the net magnetic force per meter of wirelengthon wire 4?

Equation 29-4 gives the magnitude Bof the magnetic field set up by a current in an infinitely long straight wire, at a point Pat perpendicular distance R from the wire. Suppose that point P is actually at perpendicular distance Rfrom the midpoint of a wire with a finite length L.Using Eq. 29-4 to calculate Bthen results in a certain percentage error. What value must the ratio LRexceed if the percentage error is to be less than 1.00%? That is, what LRgives

BfromEq.29-4-BactualBactual100%=1.00%?

A solenoid 1.30 long and2.60cm in diameter carries a current of 1.80A. The magnetic field inside the solenoid is 23.0mT . Find the length of the wire forming the solenoid.

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?

Figure shows a snapshot of a proton moving at velocityv=200msj^toward a long straight wire with current i=350mA. At the instant shown, the proton’s distance from the wire is d=2.89cm. In unit-vector notation, what is the magnetic force on the proton due to the current?

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.

Sign-up for free