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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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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 co
ectness. 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 ca
ies a cu
ent I. In the x-y plane there is a thin
ectangular 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.
) 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
elevant difference in the B-fields by a derivative.
Problem 3
A long solenoid of radius a, ca
ying 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 cu
ent in the solenoid is increasing at a constant rate (dI/dt = k), what cu
ent
flows in the loop, and in which way (left or right) does it pass through the resistor?
(b) If the cu
ent 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 cu
ent 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 cu
ent 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 cu
ent 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 cu
ent I flows
in the outer ring. What is the mutual inductance of the two
ings?
2
Answered Same Day Apr 05, 2021

Solution

Valupadasu answered on Apr 07 2021
146 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 cu
ent ca
ying 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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