Shown in the following figure is a long, straight wire and a single-turn rectangular loop, both of which
lie in the plane of the page. The wire is parallel to the long sides of the loop and is 0.5m away from the
closer side. At an instant when the voltage induced in the loop is 3V, what is the time rate of change of
the current in the wire?
a. What is the expression for magnetic field B due to the current I in the long, straight wire, at
distance r away from the wire?
B=
. Give your answer in terms of given variables (I,r) and
physical and numerical constants , and/or c ). Spell out Greek letters and use underscore
("-) for subscripts.
b. Because the magnetic field is not uniform, you will have to use integration to calculate the
magnetic flux, Phi =int BdA. Find the magnetic flux through the loop as a function of I ( Phi will
be proportional to I, as shown below; find the coefficient, in base SI units).
Break up the area integral, int BdA, into horizontal and vertical directions, ∬B(dx*dy). The
horizontal integral is simple (magnetic field does not change); use the expression for magnetic
field in (a) to set up the vertical integral, int B(y)dy. Set it up as a definite integral from closer
side of the loop to the far side of the loop.
Phi =
I. Give your answer in terms of given variable (I); plug in
numerical values of given geometric parameters and physical and numerical constants in base SI
units.
c. Find the rate of change of the current by setting d(Phi )/(d)t as equal to the induced voltage and
solving for d(I)/(d)t.
Note your answer in (b) has only one variable I. This makes it simple to express d(Phi )/(d)t in terms of
d(I)/(d)t.
The rate of change of the current in the wire is (dI)/(dt)=
(A)/(s).
(Note: the number here will be unreasonably large. Just plug in your result.)
Shown in the following figure is a long,straight wire and a single-turn rectangular loop, both of which lie in the plane of the page.The wire is parallel to the long sides of the loop and is 0.5 m away from the closer side.At an instant when the voltage induced in the loop is 3 V,what is the time rate of change of the current in the wire?
0.50m
0.50 m
3.0m
a.What is the expression for magnetic field B due to the current I in the long,straight wire, at distance r away from the wire?
B
Give your answer in terms of given variables I,rand
physical and numerical constants o,T, Ke and/or c. Spell out Greek letters and use underscore for subscripts.
b. Because the magnetic field is not uniform, you will have to use integration to calculate the magnetic flux,= BdA.Find the magnetic flux through the loop as a function of I will
be proportional to Ias shown below;find the coefficient,in base Sl units
Hint for (b
Break up the area integral, BdA into horizontal and vertical directions
Bdxdy.The horizontal integral is simple (magnetic field does not change); use the expression for magnetic field in (a) to set up the vertical integral, Bydy.Set it up as a definite integral from closer
side of the loop to the far side of the loop.
=
I.Give your answer in terms of given variable (I;plug in
numerical values of given geometric parameters and physical and numerical constants in base SI units.
c.Find the rate of change of the current by setting d/dt as equal to the induced voltage and solving for dI/dt.
Hint for (c Note your answer in (b)has only one variable I.This makes it simple to express d/dt in terms of dI/dt. IP A The rate of change of the current in the wire is dt S (Note: the number here will be unreasonably large. Just plug in your result.)
