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Ch 30: Electromagnetic Induction
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 30, Problem 43b

CALC An 8.0 cm×8.0 cm square loop is halfway into a magnetic field perpendicular to the plane of the loop. The loop's mass is 10 g and its resistance is 0.010 Ω. A switch is closed at t = 0 s, causing the magnetic field to increase from 0 to 1.0 T in 0.010 s. Hint: What is the impulse on the loop? With what speed is the loop 'kicked' away from the magnetic field?

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Step 1: Calculate the area of the square loop. The loop is 8.0 cm × 8.0 cm, so the area can be calculated using the formula \( A = l \times w \), where \( l \) and \( w \) are the length and width of the loop. Convert the dimensions to meters before calculating.
Step 2: Determine the change in magnetic flux through the loop. Magnetic flux \( \Phi \) is given by \( \Phi = B \times A \), where \( B \) is the magnetic field strength and \( A \) is the area of the loop. Since the magnetic field changes from 0 to 1.0 T, calculate the change in flux \( \Delta \Phi \).
Step 3: Calculate the induced electromotive force (EMF) in the loop using Faraday's law of induction. Faraday's law states \( \text{EMF} = - \frac{\Delta \Phi}{\Delta t} \), where \( \Delta t \) is the time interval over which the magnetic field changes. Substitute the values for \( \Delta \Phi \) and \( \Delta t \).
Step 4: Determine the induced current in the loop using Ohm's law. Ohm's law states \( I = \frac{\text{EMF}}{R} \), where \( R \) is the resistance of the loop. Substitute the calculated EMF and the given resistance to find the current.
Step 5: Calculate the impulse on the loop. The force on the loop is due to the interaction between the induced current and the magnetic field. The impulse \( J \) can be calculated using \( J = F \times \Delta t \), where \( F \) is the magnetic force given by \( F = I \times L \times B \), with \( L \) being the length of the side of the loop. Use the impulse-momentum theorem \( J = m \times v \) to find the speed \( v \) of the loop, where \( m \) is the mass of the loop.

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Key Concepts

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

Faraday's Law of Electromagnetic Induction

Faraday's Law states that a change in magnetic flux through a loop induces an electromotive force (EMF) in the loop. The induced EMF is proportional to the rate of change of the magnetic flux, which can be calculated using the formula EMF = -dΦ/dt, where Φ is the magnetic flux. This principle is crucial for understanding how the changing magnetic field in the problem generates an induced current in the loop.
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Lorentz Force

The Lorentz force describes the force experienced by a charged particle moving through a magnetic field. It is given by the equation F = q(v × B), where F is the force, q is the charge, v is the velocity of the particle, and B is the magnetic field. In the context of the loop, the induced current creates a magnetic field that interacts with the external magnetic field, resulting in a net force that 'kicks' the loop away.
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Impulse and Momentum

Impulse is defined as the change in momentum of an object when a force is applied over a period of time. It can be calculated using the formula Impulse = FΔt, where F is the force and Δt is the time duration. In this scenario, the impulse on the loop can be determined from the Lorentz force acting on it, leading to a change in its momentum and ultimately giving it a speed as it moves away from the magnetic field.
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Related Practice
Textbook Question

A 2.0 cm×2.0 cm square loop of wire with resistance 0.010 Ω has one edge parallel to a long straight wire. The near edge of the loop is 1.0 cm from the wire. The current in the wire is increasing at the rate of 100 A/s. What is the current in the loop?

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Textbook Question

At t = 0 s, the current in the circuit in FIGURE EX30.35 is I0. At what time in μs is the current (1/2)I0?

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Textbook Question

CALC A 10 cm×10 cm square loop of wire lies in the xy-plane. The magnetic field in this region of space is B=(0.30ti^+0.50t2k^) T\(\vec{B}\) = (0.30t\(\hat{i}\) + 0.50t^2\(\hat{k}\))\(\text{ T}\), where t is in s. What is the emf induced in the loop at (a) t = 0.5 s and (b) t = 1.0 s?

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Textbook Question

FIGURE P30.48 shows two 20-turn coils tightly wrapped on the same 2.0-cm-diameter cylinder with 1.0-mm-diameter wire. The current through coil 1 is shown in the graph. Determine the current in coil 2 at (a) t = 0.05 s and (b) t = 0.25 s. A positive current is into the figure at the top of a loop. Assume that the magnetic field of coil 1 passes entirely through coil 2.

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Textbook Question

FIGURE P30.47 shows a 1.0-cm-diameter loop with R = 0.50 Ω inside a 2.0-cm-diameter solenoid. The solenoid is 8.0 cm long, has 120 turns, and carries the current shown in the graph. A positive current is cw when seen from the left. Determine the current in the loop at t = 0.010 s.

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Textbook Question

A 100-turn, 2.0-cm-diameter coil is at rest with its axis vertical. A uniform magnetic field 60° away from vertical increases from 0.50 T to 1.50 T in 0.60 s. What is the induced emf in the coil?

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