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Ch 27: Magnetic Field and Magnetic Forces
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 27, Problem 35

A long wire carrying 4.50 A of current makes two 90° bends, as shown in Fig. E27.35. The bent part of the wire passes through a uniform 0.240 T magnetic field directed as shown in the figure and confined to a limited region of space. Find the magnitude and direction of the force that the magnetic field exerts on the wire.

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1
Identify the sections of the wire that are within the magnetic field. In the image, the wire makes two 90° bends and passes through a uniform magnetic field region. The sections of the wire within the magnetic field are the horizontal and vertical segments.
Use the formula for the magnetic force on a current-carrying wire: F = ILBsinθ, where I is the current, L is the length of the wire in the magnetic field, B is the magnetic field strength, and θ is the angle between the wire and the magnetic field.
Determine the angle θ for each segment of the wire. For the horizontal segment, the angle between the wire and the magnetic field is 90°, and for the vertical segment, the angle is 0°.
Calculate the force on each segment separately. For the horizontal segment, use F = ILB since sinθ = 1. For the vertical segment, the force is zero because sinθ = 0.
Add the forces vectorially to find the net force on the wire. Since the force on the vertical segment is zero, the net force is equal to the force on the horizontal segment. The direction of the force can be determined using the right-hand rule, which indicates that the force is perpendicular to both the current direction and the magnetic field direction.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Magnetic Force on a Current-Carrying Wire

The magnetic force on a current-carrying wire is given by the equation F = I(L × B), where I is the current, L is the length vector of the wire segment, and B is the magnetic field. The direction of the force is determined by the right-hand rule, which states that if the thumb points in the direction of the current and the fingers point in the direction of the magnetic field, the palm faces the direction of the force.
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Right-Hand Rule

The right-hand rule is a mnemonic for understanding the direction of the magnetic force on a current-carrying wire. When the thumb of the right hand points in the direction of the current and the fingers point in the direction of the magnetic field, the force exerted by the magnetic field is in the direction that the palm faces. This rule helps determine the orientation of the force vector in three-dimensional space.
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Magnetic Field Interaction

A magnetic field exerts a force on a wire carrying an electric current when the wire is within the field. The force depends on the current's magnitude, the wire's length within the field, and the angle between the wire and the magnetic field lines. In this scenario, the wire's bends and the uniform magnetic field influence the calculation of the net force, requiring consideration of each wire segment's orientation relative to the field.
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Related Practice
Textbook Question

A straight, 2.5 m wire carries a typical household current of 1.5 A (in one direction) at a location where the earth's magnetic field is 0.55 gauss from south to north. Find the magnitude and direction of the force that our planet's magnetic field exerts on this wire if it is oriented so that the current in it is running (a) from west to east, (b) vertically upward, (c) from north to south. (d) Is the magnetic force ever large enough to cause significant effects under normal household conditions?

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Textbook Question

A straight, vertical wire carries a current of 2.60 A downward in a region between the poles of a large superconducting electromagnet, where the magnetic field has magnitude B = 0.588 T and is horizontal. What are the magnitude and direction of the magnetic force on a 1.00 cm section of the wire that is in this uniform magnetic field, if the magnetic field direction is (a) east?

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Textbook Question

A 150 V battery is connected across two parallel metal plates of area 28.5 cm2 and separation 8.20 mm. A beam of alpha particles (charge +2e, mass 6.64 x 10-27 kg) is accelerated from rest through a potential difference of 1.75 kV and enters the region between the plates perpendicular to the electric field, as shown in Fig. E27.29. What magnitude and direction of magnetic field are needed so that the alpha particles emerge undeflected from between the plates?

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Textbook Question

Figure E27.49 shows a portion of a silver ribbon with z1 = 11.8 mm and y1 = 0.23 mm, carrying a current of 120 A in the +x-direction. The ribbon lies in a uniform magnetic field, in the y-direction, with magnitude 0.95 T. Apply the simplified model of the Hall effect presented in Section 27.9. If there are 5.85 x 1028 free electrons per cubic meter, find (a) the magnitude of the drift velocity of the electrons in the x-direction; (b) the magnitude and direction of the electric field in the z-direction due to the Hall effect; (c) the Hall emf.

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Textbook Question

Singly ionized (one electron removed) atoms are accelerated and then passed through a velocity selector consisting of perpendicular electric and magnetic fields. The electric field is 155 V/m and the magnetic field is 0.0315 T. The ions next enter a uniform magnetic field of magnitude 0.0175 T that is oriented perpendicular to their velocity. (a) How fast are the ions moving when they emerge from the velocity selector? (b) If the radius of the path of the ions in the second magnetic field is 17.5 cm, what is their mass?

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Textbook Question

A thin, 50.0 cm long metal bar with mass 750 g rests on, but is not attached to, two metallic supports in a uniform 0.450 T magnetic field, as shown in Fig. E27.37. A battery and a 25.0 Ω resistor in series are connected to the supports. (a) What is the highest voltage the battery can have without breaking the circuit at the supports? (b) The battery voltage has the maximum value calculated in part (a). If the resistor suddenly gets partially short-circuited, decreasing its resistance to 2.00 Ω, find the initial acceleration of the bar.

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