Where Did I Go Wrong in Calculating Net Current Through a Circle?

[hidemi]

Homework Statement

The magnetic field at any point in the xy plane is given by B vector = A r vector times K, where r vector is the position vector of the point, A is a constant, and K is a unit vector in the +z direction. The net current through a circle of radius R, in the xy plane and centered at the origin is given by:

A) π AR^2/μ0
B) 2 π AR/μ0
C) 4 π AR^3/3μ0
D) 2 π AR^2/μ0
E) π AR^2/2μ0

Ans : D

Relevant Equations

∮ B * dl ＝μ0 ＊ I

Here's what I did:
∮ B * dl ＝μ0 ＊ I
∮ AR * 2π*R ＝μ0 ＊ I
∮ 2π*AR^2 / μ0 = I
∮ 2π*AR^3 / 3μ0 = I

Where did I do wrong?

 

[haruspex]

"B vector = A r vector times K, where r vector is the position vector of the point, A is a constant, and K is a unit vector in the +z direction"
Not sure what that means. Is that $\vec B=A\vec r\times\hat k$?

 

[hidemi]

"B vector = A r vector times K, where r vector is the position vector of the point, A is a constant, and K is a unit vector in the +z direction"
Not sure what that means. Is that $\vec B=A\vec r\times\hat k$?

yes!

 

[Steve4Physics]

Here's what I did:
∮ B * dl ＝μ0 ＊ I
∮ AR * 2π*R ＝μ0 ＊ I
∮ 2π*AR^2 / μ0 = I
∮ 2π*AR^3 / 3μ0 = I

Where did I do wrong?

The mistake is in the step going from
“∮ B * dl ＝μ0 ＊ I”
to
“∮ AR * 2π*R ＝μ0 ＊ I “
Can you spot it?

Learning to use Latex for equations is not that hard. Click the 'Latex Guide' link (bottom/left of the window where you enter message). If you are posting regularly, it's worth the time/effort.

 

[hidemi]

The mistake is in the step going from
“∮ B * dl ＝μ0 ＊ I”
to
“∮ AR * 2π*R ＝μ0 ＊ I “
Can you spot it?

Learning to use Latex for equations is not that hard. Click the 'Latex Guide' link (bottom/left of the window where you enter message). If you are posting regularly, it's worth the time/effort.

Could you give me a hint? please.

 

[Steve4Physics]

Could you give me a hint? please.

Hint 1:
What are you integrating with respect to, in your expression “∮ AR * 2π*R ?

Hint 2:
Going from:
“∮ B * dl ＝μ0 ＊ I”
to:
“∮ AR * 2π*R ＝μ0 ＊ I “
your have removed the "dl" and replaced it with 2πR. What does that tell you?

If you still haven't got it, click on this spoiler:

By using 2πR you have completed the line-integral. So your second equation should simply be:
“AR * 2π*R ＝μ0 ＊ I “
without the “∮" symbol. You have completed the integration! No further integration is needed!

As an additional note, you should really be writing expressions such as $∮ \vec B \cdot \vec {dl}$ because the question is stated in terms of vectors and we are integrating the scalar (dot) product of two vectors.

 

[hidemi]

Hint 1:
What are you integrating with respect to, in your expression “∮ AR * 2π*R ?

Hint 2:
Going from:
“∮ B * dl ＝μ0 ＊ I”
to:
“∮ AR * 2π*R ＝μ0 ＊ I “
your have removed the "dl" and replaced it with 2πR. What does that tell you?

If you still haven't got it, click on this spoiler:

By using 2πR you have completed the line-integral. So your second equation should simply be:
“AR * 2π*R ＝μ0 ＊ I “
without the “∮" symbol. You have completed the integration! No further integration is needed!

As an additional note, you should really be writing expressions such as $∮ \vec B \cdot \vec {dl}$ because the question is stated in terms of vectors and we are integrating the scalar (dot) product of two vectors.

Thank you for answering my stupid question :)