Two concentric current-carrying wire loops of radius 3 cm and 9 cm lie in the same plane. The currents in the loops flow in the same directionand are equal in magnitude. The magnetic field at the common center of the loops is 50 mT. What would be the value of magnetic field at thecenter if the direction of the two currents was opposite to each other (but their value is kept constant)?Select one:○ 20 mT○ 10 mTO 15 mT○ 25 mT

Step 1: Step 2: Step 3: Step 4:

Answered: Two concentric current-carrying wire…

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 67A

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Similar questions

When the current through a circular loop is 6.0 A, the magnetic field at its center is 2.0104 T. What is the radius of the loop?
A square loop whose sides are 6.0-cm long is made with copper wire of radius 1.0 mm. If a magnetic field perpendicular to the loop is changing at a rate of 5.0 mT/s, what is the current in the loop?
The magnitude of the magnetic field 50 cm from a long, thin, straight wire is 8.0T . What is the current through the long wire?
The accompanying figure shows a cross-section of a long, hollow, cylindrical conductor of inner radius r1= 3.0 cm and outer radius r2= 5.0 cm. A 50-A current distributed uniformly over the cross-section flows into the page. Calculate the magnetic field at r = 2.0 cm. r = 4.0 cm. and r = 6.0 cm.
A toroid with a square cross section 3.0cm3.0cm has an inner radius of 25.0 cm. It is wound with 500 turns of wire, and it carries a current of 2.0 A. What is the strength of the magnetic field at the center of the square cross section?
Acircularcoiofwireofradius5.Ocmhas2Otums and carries a current of 2.0 A. The coil lies in a magnetic field of magnitude 0.50 T that is directed parallel to the plane of the coil. (a) What is the magnetic dipole moment of the coil? (b) What is the torque on the coil?
Determine the initial direction of the deflection of charged particles as they enter the magnetic fields as shown in Figure P22.2. Figure P22.2.
Figure CQ19.7 shows a coaxial cable carrying current I in its inner conductor and a return current of the same magnitude in the opposite direction in the outer conductor. The magnetic field strength at r = r0 is Find the ratio B/B0, at (a) r = 2r0 and (b) r = 4r0. Figure CQ19.7
A particle moving downward at a speed of 6.0106 m/s enters a uniform magnetic field that is horizontal and directed from east to west. (a) If the particle is deflected initially to the north in a circular arc, is its charge positive or negative? (b) If B = 0.25 T and the charge-to-mass ratio (q/m) of the particle is 40107 C/kg. what is ±e radius at the path? (c) What is the speed of the particle after c has moved in the field for 1.0105s ? for 2.0s?
An alpha-particle ( m=6.641027kg , q=3.21019C ) travels in a circular path of radius 25 cm in a uniform magnetic field of magnitude 1.5 T. (a) What is the speed of the particle? (b) What is the kinetic energy in electron-volts? (c) Through what potential difference must the particle be accelerated in order to give it this kinetic energy?
A cosmic-ray electron moves at 7.5 × 106 m/sinches perpendicular to Earth’s magnetic field at an altitude queer the field strength is 1.0 × 105T. What is the radius of the circular path the electron follows?
Why is the following situation impossible? Figure P28.46 shows an experimental technique for altering the direction of travel for a charged particle. A particle of charge q = 1.00 C and mass m = 2.00 1015 kg enters the bottom of the region of uniform magnetic field at speed = 2.00 105 m/s, with a velocity vector perpendicular to the field lines. The magnetic force on the particle causes its direction of travel to change so that it leaves the region of the magnetic field at the top traveling at an angle from its original direction. The magnetic field has magnitude B = 0.400 T and is directed out of the page. The length h of the magnetic field region is 0.110 m. An experimenter performs the technique and measures the angle at which the particles exit the top of the field. She finds that the angles of deviation are exactly as predicted. Figure P28.46

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