Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Question: In Figure, current I=56.2mAis set up in a loop having two radial lengths and two semicircles of radiia=5.72cm andb=9.36cm with a common centerP(a) What are the magnitude and (b) What are the direction (into or out of the page) of the magnetic field at P and the (c) What is the magnitude of the loop’s magnetic dipole moment? and (d) What is the direction of the loop’s magnetic dipole moment?

Short Answer

Expert verified
  1. Magnitude of magnetic field at point P is 4.97×10-7T.
  2. Direction of magnetic field at point P is into the page.
  3. Magnitude of the loop’s magnetic dipole moment is 1.06×10-3Am2.
  4. Direction of dipole moment is into the page.

Step by step solution

Achieve better grades quicker with Premium

  • Unlimited AI interaction
  • Study offline
  • Say goodbye to ads
  • Export flashcards

Over 22 million students worldwide already upgrade their learning with Vaia!

01

Identification of given data

  1. Current is56.2mA
  2. Radiusa=5.72cm
  3. Radius isb=9.36cm
02

Understanding the concept of magnetic field for semicircle

Semicircle wire generates a magnetic field that is half that of circle wire. We use the formula for magnetic field for semicircle, and the direction is given by right hand rule.

Formula:

%MathType!Translator!2!1!AMSLaTeX.tdl!AMSLaTeX!%MathType!MTEF!2!1!+-%feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbbjxAHX%garmWu51MyVXgaruavP1wzZbItLDhis9wBH5garuWqVvNCPvMCG4uz%3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj-hEeeu0xXdbb%a9frFj0-OqFfea0dXdd9vqaq-JfrVkFHe9pgea0dXdar-Jb9hs0dXd%bPYxe9vr0-vr0-vqpWqaaeaabiGaaiaacaWabeaaeaWaauaaaOqaaa%baaaaaaaaapeGaamOqaiabg2da9maalaaapaqaa8qacqaH8oqBcaWG%jbGaeqiUdehapaqaa8qacaaI0aGaeqiWdaNaamOuaaaaaaa!41C1!$B=\frac{{\muI\theta}}{{4\piR}}$%MathType!End!2!1!uncaught exception: Invalid chunk

in file: /var/www/html/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php line 68
#0 /var/www/html/integration/lib/php/Boot.class.php(769): com_wiris_plugin_impl_HttpImpl_1(Object(com_wiris_plugin_impl_HttpImpl), NULL, 'http://www.wiri...', 'Invalid chunk') #1 /var/www/html/integration/lib/haxe/Http.class.php(532): _hx_lambda->execute('Invalid chunk') #2 /var/www/html/integration/lib/php/Boot.class.php(769): haxe_Http_5(true, Object(com_wiris_plugin_impl_HttpImpl), Object(com_wiris_plugin_impl_HttpImpl), Array, Object(haxe_io_BytesOutput), true, 'Invalid chunk') #3 /var/www/html/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(30): _hx_lambda->execute('Invalid chunk') #4 /var/www/html/integration/lib/haxe/Http.class.php(444): com_wiris_plugin_impl_HttpImpl->onError('Invalid chunk') #5 /var/www/html/integration/lib/haxe/Http.class.php(458): haxe_Http->customRequest(true, Object(haxe_io_BytesOutput), Object(sys_net_Socket), NULL) #6 /var/www/html/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(43): haxe_Http->request(true) #7 /var/www/html/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(268): com_wiris_plugin_impl_HttpImpl->request(true) #8 /var/www/html/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(307): com_wiris_plugin_impl_RenderImpl->showImage('4f6d14a8e2365d3...', NULL, Object(PhpParamsProvider)) #9 /var/www/html/integration/createimage.php(17): com_wiris_plugin_impl_RenderImpl->createImage('" width="0" height="0" role="math" style="max-width: none;" localid="1662798136192">role="math" localid="1662798546354" B=μIθ4πrμ=IA

03

(a) Determining the Magnitude of the magnetic field at point P

Now magnetic field for semicircle is as follows

B1=μoIθ4πa=μoIπ4πa=μoI4a

And for radius b it is as follows

B2=μoI4b

So total magnetic field is as follows

role="math" localid="1662799565933" B=B1+B2=μoI4a+μoI4b=μoI4(1a+1b)=(4π×10-7NA-2)×(56.2×10-3A)(10.0572m+10.0936m)=4.97×10-7T

04

(b) Determining the direction of the magnetic field at point P

The direction of the magnetic field is given by the right-hand rule, and it is into the page.

05

(c) Determining the magnitude of the loop’s magnetic dipole moment

Now dipole moment is given as follows

μ=IA

Here

role="math" localid="1662799996915" A=πa2+b22=π0.0572m2+0.0936m22=0.0189m2

Now

μ=56.2×10-3A0.0189m2=1.06×10-3Am2

06

(d) Determining the direction of dipole moment

Since the direction is the same as the magnetic field, so it is into the page.

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

Figure 29-87 shows a cross section of a hollow cylindrical conductor of radii aand b, carrying a uniformly distributed currenti. (a) Show that the magnetic field magnitude B(r) for the radial distancer in the rangeb<r<ais given byB=μ0i2πra2-b2·(r2-b2)r

(b) Show that when r = a, this equation gives the magnetic field magnitude Bat the surface of a long straight wire carrying current i; when r = b, it gives zero magnetic field; and when b = 0, it gives the magnetic field inside a solid conductor of radius acarrying current i. (c) Assume that a = 2.0 cm, b = 1,8 cm, and i = 100 A, and then plot B(r) for the range 0<r<6.0cm .

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?

Show that the magnitude of the magnetic field produced at the center of a rectangular loop of wire of lengthLand widthW, carrying a current i, is

B=2μ0iπ·(L2+W2)12LW

Figure 29-34 shows three arrangements of three long straight wires carrying equal currents directly into or out of the page. (a) Rank the arrangements according to the magnitude of the net force on wire Adue to the currents in the other wires, greatest first. (b) In arrangement 3, is the angle between the net force on wire A and the dashed line equal to, less than, or more than 45°?

The current density J inside a long, solid, cylindrical wire of radius a=3.1mm is in the direction of the central axis, and its magnitude varies linearly with radial distance rfrom the axis according toJ=J0r/a, where J0=310A/m2. (a) Find the magnitude of the magnetic field at role="math" localid="1663132348934" r=0, (b) Find the magnitude of the magnetic fieldr=a/2 , and(c) Find the magnitude of the magnetic field r=a.

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