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Chapter 27: Problem 59

Two stiff parallel wires a distance \(d\) apart in a horizontal plane act as rails to support a light metal rod of mass \(m\) (perpendicular to each rail), Fig. \(27-49 .\) A magnetic field \(\overrightarrow{\mathbf{B}}\) directed vertically upward (outward in diagram), acts throughout. At \(t=0\), a constant current \(I\) begins to flow through the system. Determine the speed of the rod, which starts from rest at \(t=0\), as a function of time \((a)\) assuming no friction between the rod and the rails, and \((b)\) if the coefficient of friction is \(\mu_{\mathrm{k}} .(c)\) In which direction does the rod move, east or west, if the current through it heads north?

Short Answer

Expert verified
(a) The speed is \( v(t) = \frac{I L B}{m} t \); (b) \( v(t) = \left( \frac{I L B - \mu_k mg}{m} \right) t \); (c) The rod moves west.

Step by step solution

01

Analyze the Forces on the Rod

For part (a), when there is no friction, the only force acting on the rod is the magnetic force. Using the formula for the magnetic force, we have \( F = I L B \), where \( L \) is the length of the rod and \( B \) is the magnetic field. With the rod starting from rest, this force is the only accelerating force.
02

Relate Force to Motion of Rod

Using Newton's second law, \( F = ma \), where \( a \) is the acceleration of the rod, the acceleration can be expressed as \( a = \frac{F}{m} = \frac{I L B}{m} \). Since the rod starts from rest, the velocity \( v(t) \) of the rod at time \( t \) is obtained by integrating the acceleration with respect to time: \( v(t) = \int a \, dt = \int \frac{I L B}{m} \, dt = \frac{I L B}{m} \cdot t \).
03

Consider Frictional Forces for Part (b)

For part (b), if frictional forces are considered, the force due to friction is \( f = \mu_k mg \) opposing the motion. The net force becomes \( F_{net} = I L B - \mu_k mg \). Using Newton's second law again, \( F_{net} = ma \), we get the acceleration as \( a = \frac{I L B - \mu_k mg}{m} \). The velocity as a function of time becomes \( v(t) = \int \left( \frac{I L B - \mu_k mg}{m} \right) \, dt = \left( \frac{I L B - \mu_k mg}{m} \right) t \).
04

Determine the Direction of Motion

For part (c), using the right-hand rule for the force on current carrying wire (cross product of current and magnetic field), when the current flows north and the magnetic field is upward, the force (and thus the motion of the rod) is towards the west. This is because the force is determined by \( \vec{F} = I \vec{L} \times \vec{B} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Newton's Second Law
Newton鈥檚 second law is fundamentally about understanding how forces impact motion. It says: the acceleration of an object depends on the net force acting upon it and its mass. Simplified, this is expressed by the equation \( F = ma \). Here, \( F \) stands for force, \( m \) for mass, and \( a \) for acceleration.

In our exercise with the metal rod on rails, the magnetic force starts the rod's motion. We calculated this force as \( F = ILB \). Since this is the sole force acting initially (ignoring friction for part (a)), it becomes the driving factor in the rod's acceleration.

By substituting it into Newton's second law, you get \( a = \frac{ILB}{m} \). Notice how the rod's mass limits its acceleration: higher mass means lower acceleration for the same force. Integrating over time gives us velocity as \( v(t) = \frac{ILB}{m} \cdot t \), revealing how quickly it achieves speed proportional to time.
Friction in Physics
Friction is what often slows things down, acting opposite to the direction of movement. It happens due to interactions between surfaces, and in our exercise, it plays a major role in part (b).

For the rod, we consider kinetic friction, defined by the coefficient \( \mu_k \). This force is calculated as \( f = \mu_k mg \). Here, \( g \) is gravity's pull, and \( mg \) represents the rod's weight. When friction is present, it counteracts the magnetic force, slowing down acceleration.

This makes the net force \( F_{net} = ILB - \mu_k mg \). Using Newton鈥檚 law again, we derive \( a = \frac{ILB - \mu_k mg}{m} \) as a new acceleration. Integrating this net force, the velocity becomes \( v(t) = \left( \frac{ILB - \mu_k mg}{m} \right) t \), demonstrating how friction reduces speed.
Right-Hand Rule for Magnetic Force
The right-hand rule is a simple way to determine the direction of magnetic force on a current-carrying conductor like our rod. Imagine your right hand: point the thumb in the direction of current flow, and curl your fingers in toward your palm pointing in the magnetic field's direction. Your palm then pushes in the force's direction.

In the problem, the current flows north through the rod. The magnetic field aims upward out of the diagram.
The resulting force, due to the right-hand rule, is directed to the west. This gives insight into the rod's movement direction as it reacts to these forces.

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Most popular questions from this chapter

An electron experiences the greatest force as it travels \(2.8 \times 10^{6} \mathrm{~m} / \mathrm{s}\) in a magnetic field when it is moving northward. The force is vertically upward and of magnitude \(8.2 \times 10^{-13} \mathrm{~N}\). What is the magnitude and direction of the magnetic field?

A proton moves through a region of space where there is a magnetic field \(\overrightarrow{\mathbf{B}}=(0.45 \hat{\mathbf{i}}+0.38 \hat{\mathbf{j}}) \mathrm{T}\) and an electric field \(\overrightarrow{\mathbf{E}}=(3.0 \hat{\mathbf{i}}-4.2 \hat{\mathbf{j}}) \times 10^{3} \mathrm{~V} / \mathrm{m}\). At a given instant, the proton's velocity is \(\overrightarrow{\mathbf{v}}=(6.0 \hat{\mathbf{i}}+3.0 \hat{\mathbf{j}}-5.0 \hat{\mathbf{k}}) \times 10^{3} \mathrm{~m} / \mathrm{s}\). Determine the components of the total force on the proton.

(I) \((a)\) What is the force per meter of length on a straight wire carrying a 9.40 -A current when perpendicular to a 0.90 -T uniform magnetic field? (b) What if the angle between the wire and field is \(35.0^{\circ} ?\)

If the restoring spring of a galvanometer weakens by \(15 \%\) over the years, what current will give full-scale deflection if it originally required \(46 \mu \mathrm{A} ?\)

(II) A Hall probe used to measure magnetic field strengths consists of a rectangular slab of material (free-electron density \(n )\) with width \(d\) and thickness \(t,\) carrying a current \(I\) along its length \(\ell\) . The slab is immersed in a magnetic field of magnitude \(B\) oriented perpendicular to its rectangular face (of area \(\ell d ),\) so that a Hall emf \(\mathscr{E}_{\mathrm{H}}\) is produced across its width \(d .\) The probe's magnetic sensitivity, defined as \(K_{\mathrm{H}}=8_{\mathrm{H}} / I B,\) indicates the magnitude of the Hall emf achieved for a given applied magnetic field and current. A slab with a large \(K_{H}\) is a good candidate for use as a Hall probe. (a) Show that \(K_{H}=1 /\) ent. Thus, a good Hall probe has small values for both \(n\) and \(t\) . \((b)\) As possible candidates for the material used in a Hall probe, consider \(\left(\) i) a typical metal \(\left(n \approx 1 \times 10^{29} / \mathrm{m}^{3}\right)\) and \right. (ii) a (doped) semiconductor \(\left(n \approx 3 \times 10^{22} / \mathrm{m}^{3}\right) .\) Given that a semiconductor slab can be manufactured with a thickness of \(0.15 \mathrm{mm},\) how thin \((\mathrm{nm})\) should a metal slab be to yield a \(K_{\mathrm{H}}\) value equal to that of the semiconductor slab? Compare this metal slab thickness with the 0.3 -nm size of a typical metal atom. (c) For the typical semiconductor slab described in part \((b),\) what is the expected value for \(\mathscr{E}_{\mathrm{H}}\) when \(I=100 \mathrm{mA}\) and \(B=0.1 \mathrm{T}\) ?
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