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Ch. 27 - Magnetism
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
All textbooksGiancoli Douglas 5th editionCh. 27 - MagnetismProblem 67
Chapter 26, Problem 67

A uniform conducting rod of length ℓ and mass m sits atop a fulcrum, which is placed a distance ℓ/4 from the rod’s left-hand end and is immersed in a uniform magnetic field of magnitude B directed into the page (Fig. 27–54). An object whose mass M is 7.0 times greater than the rod’s mass is hung from the rod’s left-hand end. What current (direction and magnitude) should flow through the rod in order for it to be “balanced” (i.e., be at rest horizontally) on the fulcrum? (Flexible connecting wires which exert negligible force on the rod are not shown.)


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Identify the forces acting on the system: The rod is subject to gravitational forces due to its own weight and the weight of the hanging object, as well as the magnetic force due to the current flowing through it. The system is balanced when the net torque about the fulcrum is zero.
Write the torque equation about the fulcrum: The torque due to the weight of the rod acts at its center of mass, which is located at a distance ℓ/2 from the left-hand end. The torque due to the hanging object acts at a distance ℓ/4 from the fulcrum. The magnetic force creates a torque in the opposite direction, and its magnitude is given by the formula F = IℓB, where I is the current, ℓ is the length of the rod, and B is the magnetic field.
Set up the torque balance equation: The clockwise torques (due to the weight of the rod and the hanging object) must equal the counterclockwise torque (due to the magnetic force). Mathematically, this can be expressed as: (Mg)(ℓ/4) + (mg)(ℓ/4) = (IℓB)(ℓ/4).
Simplify the equation: Substitute M = 7m (as given in the problem) into the torque equation. This gives: (7mg)(ℓ/4) + (mg)(ℓ/4) = (IℓB)(ℓ/4).
Solve for the current I: Cancel out common terms (ℓ/4) from both sides of the equation and isolate I. The resulting expression for the current is: I = [(7mg) + (mg)] / (ℓB). This gives the magnitude of the current required to balance the system. The direction of the current can be determined using the right-hand rule for the magnetic force.

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Key Concepts

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

Torque

Torque is a measure of the rotational force acting on an object about a pivot point. It is calculated as the product of the force applied and the distance from the pivot point to the line of action of the force. In this scenario, the torques created by the weights of the rod and the hanging mass must balance each other for the rod to remain horizontal.

Magnetic Force on a Current-Carrying Conductor

When a current flows through a conductor placed in a magnetic field, it experiences a magnetic force. This force is given by the equation F = I(L × B), where I is the current, L is the length of the conductor in the field, and B is the magnetic field strength. The direction of this force can be determined using the right-hand rule, which is crucial for understanding how to balance the rod.

Equilibrium

Equilibrium in physics refers to a state where all forces and torques acting on an object are balanced, resulting in no net movement. For the rod to be at rest horizontally, the sum of the torques due to the weights and the magnetic force must equal zero. This condition allows us to determine the required current to achieve balance.
Related Practice
Textbook Question

Near the equator, the Earth’s magnetic field points almost horizontally to the north and has magnitude B = 0.50 x 10⁻⁴ T. What should be the magnitude and direction for the velocity of an electron if its weight is to be exactly balanced by the magnetic force?

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Textbook Question

Suppose the Earth’s magnetic field at the equator has magnitude 0.50 x 10⁻⁴ T and a northerly direction at all points. Estimate the speed a singly ionized uranium ion ( m = 238 u, q = e) would need to circle the Earth 5.0 km above the equator. Can you ignore gravity? [Ignore relativity.]

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Textbook Question

Suppose the electric field between the electric plates in the mass spectrometer of Fig. 27–34 is 2.84 x 10⁴ V/m and the magnetic fields are B = B'= 0.58 T. The source contains carbon isotopes of mass numbers 12, 13, and 14 from a long-dead piece of a tree. (To estimate atomic masses, multiply by 1.67 x 10⁻²⁷ kg.) Does it matter if the ion charge is positive (lost electrons) or negative (gained electrons)? Explain.

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Textbook Question

The magnetic field B at the center of a circular coil of wire carrying a current I (as in Fig. 27–9) is B = (μ₀NI) / 2r where N is the number of loops in the coil and r is its radius. Imagine a simple model in which the Earth’s magnetic field of about 1 G ( = 1 x 10⁻⁴ T) near the poles is produced by a single current loop around the equator. Roughly estimate the current this loop would carry.

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Textbook Question

A mass spectrometer is monitoring air pollutants. It is difficult, however, to separate molecules of nearly equal mass such as CO (28.0106 u) and N₂ (28.0134 u). How large a radius of curvature must a spectrometer have (Fig. 27–34) if these two molecules are to be separated on the detector by 0.50 mm?

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