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 ⃗ = ( 0.30 t i ^ + 0.50 t 2 k ^ ) 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?

Welcome back. Everyone. In this problem, a rectangular loop made of conductive material measures 15 centimeters by 20 centimeters and is oriented parallel to the xy plane. The magnetic field in this region is B equals 0.40 T I plus 0.60 T squared K teslas. And that's written in vector form. When T is 0.7 seconds, calculate the electromotive force induced within the loop for our answer choices A is 2.5 times 10 to the negative two volts. B is 4.3 times 10 to the negative six volts. C is 1.8 times 10 to the negative four volts and D is 6.9 times 10 to the negative three volts. So let's first make a sketch of what's really going on here. So we have a rectangular loop probably looks something like this. OK? And we know that our rectangular loop measures 15 centimeters by 20 centimeters. OK. And we also know I treat it like this because we also know it's oriented parallel to the xy plane. So if I were to draw my xy plane, just an idea here, then we know it would look something like this. OK. So we have our xy. So that's why we have our victor that's written in terms of I and K. OK? Because that's our xy plane. No, we want to find the electromotive force that's induced within the loop when T is 0.7 seconds. In other words, we want to find that induced voltage after 0.7 seconds. And if we're going to find it and we're dealing with a vector, then we need to think about what links our electromotive force to any information that we have on vectors for magnetic flux or for the magnetic field. Rather. Well, let's think about it. What links electromotive force to a change in magnetic field? Well, recall that are electromotive force is or is induced by a change in flux over time. So this is what we know. In other words, the induced EMF is the derivative of flux over time. OK. Now there's no information here that tells us about the flux, but we know that the magnetic flux through a surface is the dot product of the magnetic field vector and the area vector of the surface. In other words, we know that our magnetic flux with respect to time is the dot product of the K component of the magnetic field multiplied by the area of vector because the area of vector is perpendicular to the plane of the loop in the K direction. Since we have the magnetic field vector component in the K direction. If we can figure out the area, then we can figure out the flux with respect to, we can figure out the flux as a function of time and then differentiate it with respect to time when the time equals 0.7 seconds. So let's go ahead and do that. The care component of our magnetic field is 0.60 T squared. So that means our magnetic flux with respect to time will be equal to 0.60 T squared multiplied by the area of our rectangular loop which would be 15 centimeters multiplied by 20 centimeters. So that's zero point 1 5 m multiplied by 0.2 m. OK. That would be quite true. 0.03 square meters. And the product of those two would be equal to 0.0 and eight T squared tesla meters squared because I remember that was in teslas. OK. So what does this mean for us? Well, now that we have the flux with respect to time, we can differentiate it because now that tells us then that the derivative of or a flux with respect to time. Remember when we differentiate a variable, we bring down the power and multiply it by our term and then subtract one from the power. So it's gonna be two multiplied by 0.018 T is the power of two minus one. OK. And when we differentiate that, that will be equal to 0.036 T votes. So that is our electromotive force written as magnetic flocks with respect to time. Therefore, at T equals 0.7 seconds, right, our electromotive force is going to be equal to 0.036 multiplied by seven. And that equals 0.025 volts written in standard form that 2.5 times 10 to the negative two volts. Therefore, an electromotive force of 2.5 times 10 to the negative two volts is induced after 0.7 seconds. When we look back on our answer choices that would have been answer choice. A thanks a lot for watching everyone. I hope this video helps.

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30. Induction and Inductance

Faraday's Law

Physics

30. Induction and Inductance

Faraday's Law

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Problem 40

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 $\vec{B} = (0.30 t \hat{i} + 0.50 t^{2} \hat{k}) T$ , 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?

Verified step by step guidance

Understand the problem: The problem involves calculating the electromotive force (emf) induced in a square loop of wire due to a time-varying magnetic field. The emf is related to the rate of change of magnetic flux through the loop.

Write the formula for emf: The induced emf (ε) is given by Faraday's Law of Induction: ε = -dΦ/dt, where Φ is the magnetic flux through the loop. Magnetic flux is defined as Φ = ∫B⃗ ·dA⃗, where B⃗ is the magnetic field and dA⃗ is the area vector perpendicular to the loop.

Determine the magnetic flux: The loop lies in the xy-plane, so the area vector dA⃗ points in the z-direction (k̂). The magnetic field component contributing to the flux is the z-component, Bz = 0.50t². The area of the square loop is A = (0.10 m)² = 0.01 m². Thus, Φ = Bz × A = (0.50t²) × (0.01).

Differentiate the flux with respect to time: To find the emf, calculate dΦ/dt. Differentiate Φ = 0.50t² × 0.01 with respect to t. This gives dΦ/dt = d/dt [0.50 × 0.01 × t²] = 0.50 × 0.01 × 2t = 0.01t.

Substitute the given times: For part (a), substitute t = 0.5 s into the expression for dΦ/dt to find the emf. For part (b), substitute t = 1.0 s into the same expression. Remember to include the negative sign from Faraday's Law, so ε = -0.01t.

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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 closed loop induces an electromotive force (emf) in the loop. The induced emf is proportional to the rate of change of magnetic flux, which can be calculated using the formula emf = -dΦ/dt, where Φ is the magnetic flux. This principle is fundamental in understanding how electric currents can be generated by changing magnetic fields.

Magnetic Flux

Magnetic flux (Φ) is a measure of the quantity of magnetism, taking into account the strength and the extent of a magnetic field. It is defined as the product of the magnetic field (B) and the area (A) through which the field lines pass, adjusted for the angle (θ) between the field lines and the normal to the surface: Φ = B·A·cos(θ). In this problem, the magnetic field varies with time, affecting the flux through the loop.

Induced EMF Calculation

To calculate the induced emf in the loop at specific times, one must first determine the magnetic field at those times and then compute the magnetic flux through the loop. The emf can be found by differentiating the magnetic flux with respect to time. This involves substituting the values of the magnetic field into the flux equation and applying the appropriate time derivatives to find the induced emf at t = 0.5 s and t = 1.0 s.

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CALC A 10 cm×10 cm square loop of wire lies in the xy-plane. The magnetic field in this region of space is B⃗=(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?

Key Concepts

Faraday's Law of Electromagnetic Induction

Magnetic Flux

Induced EMF Calculation

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CALC A 10 cm×10 cm square loop of wire lies in the xy-plane. The magnetic field in this region of space is $\vec{B} = (0.30 t \hat{i} + 0.50 t^{2} \hat{k}) T$ , 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?