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Chapter 29: Problem 1

A solenoid with 200 turns and a cross-sectional area of \(60 \mathrm{~cm}^{2}\) has a magnetic field of \(0.60 \mathrm{~T}\) along its axis. If the field is confined within the solenoid and changes at a rate of \(0.20 \mathrm{~T} / \mathrm{s}\), the magnitude of the induced potential difference in the solenoid will be a) \(0.0020 \mathrm{~V}\). b) \(0.02 \mathrm{~V}\). c) \(0.001 \mathrm{~V}\). d) \(0.24 \mathrm{~V}\).

Short Answer

Expert verified
Answer: To find the magnitude of the induced potential difference, we follow these steps: 1. Convert the cross-sectional area to SI units: \(60 \mathrm{~cm}^2 = 60 \times 10^{-4} \mathrm{~m}^2\). 2. Calculate the magnetic flux: \(\Phi = 200 \times 0.60 \times 60 \times 10^{-4}\). 3. Calculate the magnitude of the induced potential difference (emf) using Faraday's law: \(E = 200 \times 60 \times 10^{-4} \times 0.20\). 4. Compare the calculated induced potential difference to the given options to find the correct answer.

Step by step solution

01

Convert the cross-sectional area to SI units.

The cross-sectional area is given in cm², so convert it to m²: \(60 \mathrm{~cm}^2 = 60 \times 10^{-4} \mathrm{~m}^2\).
02

Calculate the magnetic flux.

According to Faraday's law, we need to find the magnetic flux, which can be calculated using the formula \(\Phi = NBA\), where N is the number of turns, B is the magnetic field, and A is the cross-sectional area. We can plug in the values: \(\Phi = 200 \times 0.60 \times 60 \times 10^{-4}\).
03

Calculate the magnitude of the induced potential difference (emf).

Faraday's law states that the magnitude of induced emf \(E\) is equal to the rate of change of magnetic flux \(\frac{d\Phi}{dt}\). Since the rate of change of the magnetic field is 0.20 T/s, we can calculate the rate of change of magnetic flux: \(\frac{d\Phi}{dt} = N \times A \times \frac{dB}{dt}\). Now, plugging in the given values, \(E = 200 \times 60 \times 10^{-4} \times 0.20\).
04

Determine the correct answer.

After calculating the value of the induced potential difference in the solenoid, compare it to the given options to find the correct answer. After calculating step by step, the final answer should match one of the given options.

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Most popular questions from this chapter

A square conducting loop with sides of length \(L\) is rotating at a constant angular speed, \(\omega\), in a uniform magnetic field of magnitude \(B\). At time \(t=0,\) the loop is oriented so that the direction normal to the loop is aligned with the magnetic field. Find an expression for the potential difference induced in the loop as a function of time.

A wedding ring is tossed into the air and given a spin, resulting in an angular velocity of 13.5 rev/s. The rotation axis is a diameter of the ring. The magnitude of the Earth's magnetic field is \(4.97 \cdot 10^{-5} \mathrm{~T}\) at the ring's location. If the maximum induced voltage in the ring is \(1.446 \cdot 10^{-6} \mathrm{~V},\) what is the diameter of the ring?

An ideal battery (with no internal resistance) supplies \(V_{\text {emf }}\) and is connected to a superconducting (no resistance!) coil of inductance \(L\) at time \(t=0 .\) Find the current in the coil as a function of time, \(i(t) .\) Assume that all connections also have zero resistance.

An elastic circular conducting loop expands at a constant rate over time such that its radius is given by \(r(t)=r_{0}+v t,\) where \(r_{0}=0.100 \mathrm{~m}\) and \(v=0.0150 \mathrm{~m} / \mathrm{s}\). The loop has a constant resistance of \(R=12.0 \Omega\) and is placed in a uniform magnetic field of magnitude \(B_{0}=0.750 \mathrm{~T}\), perpendicular to the plane of the loop, as shown in the figure. Calculate the direction and the magnitude of the induced current, \(i\) at \(t=5.00 \mathrm{~s}\).

The long, straight wire in the figure has a current \(i=1.00 \mathrm{~A}\) flowing in it. A square loop with 10.0 -cm sides and a resistance of \(0.0200 \Omega\) is positioned \(10.0 \mathrm{~cm}\) away from the wire. The loop is then moved in the positive \(x\) -direction with a speed \(v=10.0 \mathrm{~cm} / \mathrm{s}\) a) Find the direction of the induced current in the loop. b) Identify the directions of the magnetic forces acting on all sides of the square loop. c) Calculate the direction and the magnitude of the net force acting on the loop at the instant it starts to move.
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