A straight wire in a magnetic field experiences a force of 0.032 N when the current in the wire is 1.5 A. The current in the wire is changed, and the wire experiences a force of 0.074 N as a result. What is the new current?

A straight wire in a magnetic field experiences a force of 0.032 N when the current in the wire is 1.5 A. The current in the wire is changed, and the wire experiences a force of 0.074 N as a result. What is the new current?

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter31: Gauss’s Law For Magnetism And Ampère’s Law

Section: Chapter Questions

Problem 39PQ

See similar textbooks

Related questions

Q: A long straight wire carries a current of 51.3 A. An electron, traveling at 1.00 × 107 m/s, is 4.70…

A: Given that: The current: I=5.13 A The velocity: v=1.0×107 m/s The distance: r=4.70 cmr=4.70×10-2 m…

Q: See the attached figure for this problem. A charge of 8.55 x 10^6 C moves to the right as shown…

A: Here, qE=qvBv=EB

Q: Suppose you wish to measure the current in a wire. You have a device that can measure the strength…

Q: maximum energy that can be transferred

Q: A 1.0-m-long, 1.0-mm-diameter copper wire carries a current of 50.0 A to the east. Suppose we create…

Q: See the attached figure for this problem. A charge of 9.93 x 10^6 C moves to the right as shown…

A: Since the charge is moving at a constant velocity, the net force on the charge will be zero.Write…

Q: What is the current that is flowing through a wire that has a magnetic field strength of 8.2 x 10^-4…

A: Given that Magnetic field due to wire B = 8.2 X 10-4 T Distance from the wire r = 0.005 m

Q: What is the force per meter (in N/m) on a lightning bolt at the equator that carries 19,040 A…

A: lightning bolt at the equator carries 19040A current.Earth's magnetic field is 3.00 ✕ 10−5 T.

Q: Please refer again to the diagram. This time we have a charge of 5.52 x 10^-6 C moving at velocity…

A: Given data: Charge (q) = 5.52×10-6 CVelocity (v) = 5.73×105 m/sMagnetic field (B) = 1.53 TElectric…

Q: A straight wire that is 0.60 m long is carrying a current of 2.0 A. It is placed in a uniform…

A: Given Wire length l = 0.60 m Current I = 2.0 A Magnetic field B = 0.30 T Force F…

Q: An electron moving with a velocity 4.9 x10° m/s in a magnetic field B= 5.7 x1oSTwill move in a…

Q: A 5.47 C point charge is moving with a speed of 6.98 m/s. It is moving perpendicular to a magnetic…

A: Charge (q) = 5.47 C Speed (v) = 6.98 m/s Magnetic field (B) = 6.28 T

Q: An iron bar passes through a magnetic field at a speed of 6.3 m/s. If the bar is 1.5 m long and…

A: Given that Length of the conducting bar l = 1.5m Maximum emf ( potential difference) across it's end…

Q: A 1.5 m wire carries a current of 0.75 A traveling east. The wire is in the presence of a 0.33 T…

Q: A proton moving at a speed of 4.2 × 106 m/s cuts across the lines of a magnetic field at an angle of…

Q: See the attached figure for this problem. There is a charge of 6.24 x 10^-6 C moving at velocity…

A: Calculate the magnetic field due to the charge. Fm=qvBFm=(6.24×10−6 C)(8.62×105 m/s)(3 T)Fm=16.1 N

Q: An electron accelerated accelerated from rest through a potential difference of 4000 volts enters…

A: Magnetic field.

Q: A negatively charged ion with an excess electron moving with a velocity vx = 1.095 x103 m/s is…

Q: magnitude of the current in the wire

Q: A wire carrying a 25 A current passes between the poles of a strong magnet such that the wire is…

A: The current in the wire is 25 A. The magnetic force is 2.33 N. The length of the wire is 6 cm = 0.06…

Q: A wire carrying a 47.5 A current passes between the poles of a strong magnet so that it is…

Q: An electron with a speed of 1.2 x 102 m/s enters a magnetic field of 0.0020 T at 90 degrees to the…

A: The objective of the question is to determine the instantaneous acceleration of an electron as it…

Q: long straight wire carries a current of 49.2 A. An electron, traveling at 1.00 × 107 m/s, is 8.00 cm…

Q: a wire 31.7 m long is in a 4.55T magnetic field. A 3.65 N force acts on the wire. What current is in…

Q: Two long, parallel wires are hung by cords of length 6.0cm. Each wire has a mass per unit length of…

Q: A wire carrying a 28.0 A current passes between the poles of a strong magnet such that the wire is…

A: Given: The current in the wire is 28 A. The magnetic force is 2.25 N. The length of the wire is 6…

Q: A wire levitates in a magnetic field while a 2.5 A current passes through it. The wire's mass is 1…

A: when wire leviates magnetic force and gravitational force balance each other Fb= magnetic force =…

Q: A helium nucleus (2 protons, 2 neutrons) is traveling at a velocity of 5.27 x 10^6 m/s a distance of…

A: Velocity of the helium nucleus is Current in the wire is Distance from the wire is The helium…

Q: Two long parallel wires are a distance d apart. The current in one wire is 10 A and the current in…

Q: A wire extends from x = - (infinity) to x = 0 on the x-axis, then turns and extends from y = 0 to y…

Q: A 0.55-meter-long wire carries a current with electron flow directed to the East. It is present in a…

A: Given values: Length of the wire, l=0.55 m Intensity of magnetic field, B=0.050 T Force acting on…

Q: An electron in a TV CRT moves with a speed of 5.5 × 107 m/s, in a direction perpendicular to the…

Concept explainers

Magnetic Field Of Coaxial Cable

The word coaxial means same axis. So for a long straight cable to be coaxial, it must have concentric cylindrical shaped conductor around it. Coaxial cable (popularly called ‘coax’) has various applications ranging from current conductors to radiofrequency signal transmitters. This cable has lower emission losses and provides protection from electromagnetic interference which allows signals with less power to transmit over longer distances.For example, currents produced in the pickup of a stereo turntable generally travel to the amplifier through a coaxial cable.The self-inductance of a coaxial cable is another important characteristic, because it influences the propagation of electrical signals in the cable.

Question

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions

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?
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?
Solenoid A has length L and N turns, solenoid B has length 2L and N turns, and solenoid C has length L/2 and 2N turns. If each solenoid carries the same current, rank the magnitudes of the magnetic fields in the centers of the solenoids from largest to smallest.
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?
An electron moving with a velocity v=(4.0i+3.0j+2.0k)106m/s enters a region where there is a uniform electric field and a uniform magnetic field. The magnetic field is given by v=(1.0i2.0j+4.0k)102T. If the electron travels through a region without being deflected, what is the electric field?
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.
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.
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
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
Electrons in Earths upper atmosphere have typical speeds near 6.00 105 m/s. (a) Calculate the magnitude of Earths magnetic field if an electrons velocity is perpendicular to the magnetic field and its circular path has a radius of 7.00 102 m. (b) Calculate the number of times per second that an electron circles around a magnetic field line.
A wire 2.80 m in length carries a current of 5.00 A in a region where a uniform magnetic field has a magnitude of 0.390 T. Calculate the magnitude of the magnetic force on the wire assuming the angle between the magnetic field and the current is (a) 60.0, (b) 90.0, and (c) 120.