Electromagnetic Induction exercise, is something wrong with these answers?

[Stephen Bulking]

Homework Statement

This is not actually my homework since I already have the answers but the explanation is a little confusing, please check.
1. The resistance of the rectangular current loop is R. The metal rod is sliding to the right with the constant velocity of v. An infinite current I is placed near the loop(see figure). (please see word file attached)

The induced current around the loop:
answer is Counterclockwise , i=(μ0 vI×ln(b ∕ a))/2πR

3.Calculate the self-induced emf in the solenoid if the current it carries is decreasing at the rate of 50 A/s, know that the inductance of an air-core solenoid containing 300 turns and the length of the solenoid is 25 cm and its cross-sectional is 4 cm2:
A.-8.05mV B.5.30 mV C.13.05 mV D. 9.05mV

Relevant Equations

Lens' law
ε= -dΦ∕dt= -Ldi/dt
|ε|= dΦ∕dt
dΦ=Bds
ΔV=ε
i: induced current
L:inductant

1.
|ε|=dΦ/dt
B=μ0I×cos(0-cos180)/4πr
=μ0I/2πr
dΦ=BdS=μ0I×(vdtxdr)/2πr
ΔΦ=(μ0I×(vdt)/2π)x∫dr/r=(μ0I×(vdt)/2π)xln(b/a)
*the intergral goes from a to b
|ε|=dΦ/dt=(μ0I×v×ln(b/a)/2π)
i=ε/R=(μ0I×v×ln(b/a)/2πR)
2.
dI/dt=-50
B=μ0NI/L
ε=-dΦ/dt=-BdS/dt (*)
=-NBS/dt (wait what?)
=-N×N×I×S×μ0/dt
=-(N×N×S×μ0)×dI/dt (big what the hell)
= 9.05(mV)(*) my teacher wrote -∫BdS/dt but this is inconsistent with the given formula so I made this minor change

 

[BvU]

This is not actually my homework since I already have the answers

Ha ! Funny criterion to distinguish homework from other stuff.

but the explanation is a little confusing

Don't see no explanation, just a sequence of algebraic manipulations. What is the confusion ? (let's do 1. first)

 

[kuruman]

Homework Statement:: This is not actually my homework since I already have the answers but the explanation is a little confusing, please check.
1. The resistance of the rectangular current loop is R. The metal rod is sliding to the right with the constant velocity of v. An infinite current I is placed near the loop(see figure). (please see word file attached)

The induced current around the loop:
answer is Counterclockwise , i=(μ0 vI×ln(b ∕ a))/2πR

3.Calculate the self-induced emf in the solenoid if the current it carries is decreasing at the rate of 50 A/s, know that the inductance of an air-core solenoid containing 300 turns and the length of the solenoid is 25 cm and its cross-sectional is 4 cm2:
A.-8.05mV B.5.30 mV C.13.05 mV D. 9.05mV
Homework Equations:: Lens' law
ε= -dΦ∕dt= -Ldi/dt
|ε|= dΦ∕dt
dΦ=Bds
ΔV=ε
i: induced current
L:inductant

1.
|ε|=dΦ/dt
B=μ0I×cos(0-cos180)/4πr
=μ0I/2πr
dΦ=BdS=μ0I×(vdtxdr)/2πr
ΔΦ=(μ0I×(vdt)/2π)x∫dr/r=(μ0I×(vdt)/2π)xln(b/a)
*the intergral goes from a to b
|ε|=dΦ/dt=(μ0I×v×ln(b/a)/2π)
i=ε/R=(μ0I×v×ln(b/a)/2πR)
2.
dI/dt=-50
B=μ0NI/L
ε=-dΦ/dt=-BdS/dt (*)
=-NBS/dt (wait what?)
=-N×N×I×S×μ0/dt
=-(N×N×S×μ0)×dI/dt (big what the hell)
= 9.05(mV)(*) my teacher wrote -∫BdS/dt but this is inconsistent with the given formula so I made this minor change

1. is OK. For 2 (or is it 3?) look up the expression for the self-inductance ##L## of a solenoid and use ##V_L=-L\frac{dI}{dt}##, it's that simple.

On edit: Your derivation went astray at the "wait what?" Note that ##\dfrac{d \phi}{dt}=\dfrac{d \phi}{dI}\dfrac{d I}{dt}##. Now ##\dfrac{d \phi}{dI}## is a constant that depends only on the geometry since ##B## is always proportional to ##I## according to Biot-Savart. This constant is the self inductance ##L## which you can derive separately for your geometry.

 

[rude man]

"Infinite current"? "Solenoid" - where?
My head spins ...

 

[kuruman]

"Infinite current"? "Solenoid" - where?
My head spins ...

It's an infinite current carrying wire as the picture suggests. The "carrying wire" part was omitted for some reason.

The solenoid thing appears to be an unrelated problem.

 

[Stephen Bulking]

Ha ! Funny criterion to distinguish homework from other stuff.
Don't see no explanation, just a sequence of algebraic manipulations. What is the confusion ? (let's do 1. first)

Alright so for the first one, and I'm going to write all of this in steps so you could track it more easily;
Step 1: establish that |ε|=dΦ/dt and i=ε/R which means we only need to define ε.
|ε| is emf
Φ is magnetic flux which is equal to BdS
Step 2: Find B; which is given by B=μ0I×cos(0-cos180)/4πr which is simplified to μ0I/2πr
Step 3: Which means
dΦ= BdS = μ0I×(vdtxdr)/2πr
(vdtxdr) is supposed to be the area dS but I don't get why there is dr...vdt is totally fine.
Step 4: after step 3 everything is natural and we have
ΔΦ = μ0I×(vdt)/2π)xln(b/a)
The integral of course goes from a to b
Step 5:Plug everything in and hope the answer looks like one of the answers
|ε|=dΦ/dt=(μ0I×v×ln(b/a)/2π)

 

[Stephen Bulking]

1. is OK. For 2 (or is it 3?) look up the expression for the self-inductance ##L## of a solenoid and use ##V_L=-L\frac{dI}{dt}##, it's that simple.

On edit: Your derivation went astray at the "wait what?" Note that ##\dfrac{d \phi}{dt}=\dfrac{d \phi}{dI}\dfrac{d I}{dt}##. Now ##\dfrac{d \phi}{dI}## is a constant that depends only on the geometry since ##B## is always proportional to ##I## according to Biot-Savart. This constant is the self inductance ##L## which you can derive separately for your geometry.

Ok I totally agree and have solved the problem, thank you. But what do you mean by "This constant is the self inductance ##L## which you can derive separately for your geometry." and also if able, please check my other comment to BvU on question 1, I still have a little trouble on that one.

 

[BvU]

Alright so for the first one, and I'm going to write all of this in steps so you could track it more easily;
Step 1: establish that |ε|=dΦ/dt and i=ε/R which means we only need to define ε.
|ε| is emf
Φ is magnetic flux which is equal to BdS
Step 2: Find B; which is given by B=μ0I×cos(0-cos180)/4πr which is simplified to μ0I/2πr
Step 3: Which means
dΦ= BdS = μ0I×(vdtxdr)/2πr
(vdtxdr) is supposed to be the area dS but I don't get why there is dr...vdt is totally fine.
Step 4: after step 3 ...

For step 1 I don't see why you change over from ##\varepsilon## to ##|\varepsilon|## ?
That way, with ##\varepsilon = -{d\Phi\over dt}## you lose the sign

In Step 2 you implicitly choose coordinates (or do you mean scalar multiplication with your ##\times## sign ? And then determine the direction of ##\vec B## separately ? How ?

It is better to choose a (right handed) coordinate system first and then use Biot-Savart to determine the direction of ##B##. You don't have to carry out the integration, that's been done already and gets ##\mu_0\over 2\pi r##. Either that, or you use Ampere and somehow know the direction of ##\vec B##. Result: With ##\vec I## to the left, ##\vec B## is into the paper.

Next substep (still in 2) is the inner product $$\vec d\Phi = \vec B \cdot \vec {dS} .$$
Vector $$\vec {dS} = {\bf i} dx \times {\bf j} dy = {\bf i} \times {\bf j} (v\;dt \, dy) = {\bf k} (v\;dt \, dy) $$
So that [edit:] corrected in #10 $$\vec d\Phi = \vec B \cdot \vec {dS} = - {\mu_0\over 2\pi }y \;{\bf k} \cdot {\bf k} (v\;dt \, dy) = - {\mu_0\over 2\pi y} (v\;dt \, dy) $$ and if ##{d\Phi} = a\, dt## then ##{d\Phi\over dt} = a##, which leaves you with an expression in ##dy##, ##y## is running from a to b.

 

[Stephen Bulking]

For step 2, that is the scalar product and as for the direction, I used the right-hand-rule to determined the GIVEN magnetic field's direction and since there's change in the magnetic flux as the bar thing moves to the right, there is an INDUCED current whose magnetic field is opposing the change in magnetic flux. (Lenz's law I believe).
Also for step 2 (extension), I'm having trouble with your math:
So on your "Vector line" you group dx and dy together, just a stupid question but this is possible right?...And then you straight up cross product that thing, nice, nice...
And then on your "So that line" you turned the 1/r into yk... how? oh wait you just changed the notation from r to y, ok that's cool and you just put y on the numerator...typo?
But what I NEED to understand is why is there dr or dy in the first place; it's a bar and it moves in a swiping motion left to right, it's length doesn't change so why is there dr or dy??

 

[BvU]

my teacher wrote -∫BdS/dt

which is correct and consistent.

but this is inconsistent with the given formula

Don't see no given formula in the problem statement ?

the GIVEN magnetic field's direction

Don't see that in the problem statement either ?

just a stupid question but this is possible right?

Not a stupid question and yes, the orher factors are constants and one can use ab = ba

you turned the 1/r into yk... how?

1/r is 1/y -- but I misplaced a curly bracket in the ##\LaTeX##. It should be:

$$ d\Phi = \vec B \cdot \vec {dS} = - {\mu_0\over 2\pi y} \;{\bf k} \cdot {\bf k} (v\;dt \, dy) = - {\mu_0\over 2\pi y} (v\;dt \, dy) $$

But what I NEED to understand is why is there dr or dy in the first place; it's a bar and it moves in a swiping motion left to right, it's length doesn't change so why is there dr or dy??

Kudos for picking out the most important issue here: in this exercise ##\Phi = \int \vec B \cdot \vec {dS}## changes with time, not because of ##\vec B## but because of ##\vec S## increasing with time. Write ##\vec {dS} = \vec {dx} \times \vec{dy}## and there you are.

By the way, all this stuff stems from the Lorentz force, with which you can obtain the same result by just looking at the moving bar.

 

[kuruman]

Ok I totally agree and have solved the problem, thank you. But what do you mean by "This constant is the self inductance ##L## which you can derive separately for your geometry." and also if able, please check my other comment to BvU on question 1, I still have a little trouble on that one.

The inductance ##L## is a constant that depends on the geometry, e.g. cross-section, length and number of turns of the solenoid, but not on the current. See here for a derivation that might be informative. It allows you to calculate the back emf by knowing the rate of change of the current.$$\text{emf}=-\frac{d \Phi}{dt}=-\frac{d \Phi}{dI}\frac{dI}{dt}$$With the definition of inductance ##L \equiv \dfrac{d \Phi}{dI}##,##~\text{emf}=-L\dfrac{dI}{dt}.##

 

[kuruman]

$$\vec d\Phi = \vec B \cdot \vec {dS} = \dots$$

Minor point: In this and other posts above did you mean $$d\Phi = \vec B \cdot d\vec {S} = \dots ~?$$