PHY250 Problem Set 5, 2020 Assigned: Sunday March 22, 2020. Due: Sunday April 5, 2020 (by 11:59 pm). Only Problems 1 and 3 will be marked in detail for correctness. These two problems are each worth...

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Intro to electrodynamics


PHY250 Problem Set 5, 2020 Assigned: Sunday March 22, 2020. Due: Sunday April 5, 2020 (by 11:59 pm). Only Problems 1 and 3 will be marked in detail for correctness. These two problems are each worth 35% of the total problem set marks. For each of Problems 2, 4, and 5, you will get 10% of the total marks for a significant attempt at solving the problems. For each problem, please list the individuals with whom you collaborated in solving the problem. If you collaborated with no one, write ”Collaborators: None.” Problem 1 Consider a circular wire loop in the x-y plane in the presence of a spatially uniform magnetic field B = B0(1 + kt) ẑ, where B0 and k are constants and t is time. If the loop is heated such that its radius r is changing linearly in time as r = vt, where v is the constant radial velocity of a point on the loop, determine the induced emf in the wire loop. Comment on the expression that you derived for the emf. Problem 2 A long straight wire is parallel to the y axis and is located at a distance z = h on the z axis. Assume that the wire is stationary and carries a current I. In the x-y plane there is a thin rectangular loop. The sides of the loop that are parallel to the wire have length ` and the other sides are very small (and have length b). The loop slides with constant speed v in the x̂ direction. a) Find the magnitude of the emf induced in the loop at the moment when the center of the loop is at position x. b) For what values of x does this emf have a local maximum or minimum? Hint: For this problem, work in the approximation b � x, so that you can approximate the relevant difference in the B-fields by a derivative. Problem 3 A long solenoid of radius a, carrying n turns per unit length, is looped by a wire with resis- tance R, as shown in Figure 1. 1 Figure 1 (a) If the current in the solenoid is increasing at a constant rate (dI/dt = k), what current flows in the loop, and in which way (left or right) does it pass through the resistor? (b) If the current I in the solenoid is constant but the solenoid is pulled out of the loop (toward the left, to a place far from the loop), what total charge passes through the resistor? Problem 4 Figure 2 A small loop of wire (with radius a and resistance R) is located a distance above a large loop (with ra- dius b and current I), as shown in Figure 2. The planes of the two loops are parallel, and perpendicu- lar to the common axis. As shown, the small loop is at a distance z above the large loop and moving in the z-direction with v = dz/dt > 0. Assume that z � b, so that the field of the large loop is es- sentially constant across the area bounded by the small loop. (a) Determine the flux Φ through the area of the small loop when the loop is at a distance z above the large loop. (b) Determine the emf and induced current in the small loop at the instant it is at a distance z and moving with v = v ẑ. What is the direction of the induced current in the small loop? Figure 3 Problem 5 Consider two concentric wire rings, centered on the origin, as shown in Figure 3. The inner ring has a radius Ra and the outer ring has a radius Rb, with Rb � Ra. A current I flows in the outer ring. What is the mutual inductance of the two rings? 2
Answered Same DayApr 05, 2021

Answer To: PHY250 Problem Set 5, 2020 Assigned: Sunday March 22, 2020. Due: Sunday April 5, 2020 (by 11:59 pm)....

Valupadasu answered on Apr 07 2021
133 Votes
1. B = B0(1+kt).?⃗
Area of loop ? = ??2 ?⃗: r=vt A = π v 2t 2 ?⃗
Flux= ?⃗⃗.?⃗ = B0(1+kt). π
v 2t 2
Φ = π= B0v 2( t 2 + k t 3)
Emf = |dφ/dt |
= πB0v 2(2t+3k t 2)
    Emf is a quadratic function of time


2.
a) The emf induced due to change in magnetic flux is
    Given by e= -dφ/dt = -d(BA)/dt ……….1
And it’s magnit√?2+?2ude is e= d(BA)/dt…………2
Area of loop is not changing, A=lb e= AdB/dt = lb dB/dt…………3 the magnetic field due to current carrying wire at distance r is
B = I/2πr
B = I/2π (
And x= vt
B = I/
Eq 3 becomes e = lbI/2 π d/dt of 1/
after simplifying we have e = lbI v 2 t /2π(v 2 t 2 +h 2 ) 3/2 ……………4
now...
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