Magnetic Field-Induced square loop

[dominatesean]

Homework Statement

A 11cm by 11cm square loop lies in the xy-plane. The magnetic field in this region of space is B = (.31t(i(hat))+.55t^2(k(hat)))T, t is in s.

What is the E induced in the loop at t = 0.5s?
at t = 1.0s?


2.Equations
E=d$$\phi$$_m/dt aka Faraday's Law
(greek O with an I inside)

The Attempt at a Solution

I have to convert cm into m, so it's .0121m²*(.31t+.55t²) plug in .5 and 1 into t
Wrong
tried .0121m²*(.31+1.1t)plug for t
Wrong

A little help would be appreciated...Thanks for looking at it.

 

[Doc Al]

I have to convert cm into m, so it's .0121m²*(.31t+.55t²) plug in .5 and 1 into t
Wrong

Two mistakes:
(1) You didn't find Φ properly (What's the component of B perpendicular to the loop?)
(2) You didn't find dΦ/dt

tried .0121m²*(.31+1.1t)plug for t
Wrong

See mistake (1) above.

 

[dominatesean]

Two mistakes:
(1) You didn't find Φ properly (What's the component of B perpendicular to the loop?)
(2) You didn't find dΦ/dt

Sometime physics confounds me, this would be one of those times.

Φ would that be A*B as A=the area$$\Pi$$r² and B is the field.
Then dΦ/dt would be the derivative of the change in field/change in the time...I didn't think the field was changing?

 

[Doc Al]

Φ would that be A*B as A=the area$$\Pi$$r² and B is the field.

Almost. Φ = A*B, where A is the area vector (normal to the loop) and B is the field vector and * is the scalar product (equivalent to what I said earlier about finding the component of B perpendicular to the loop).

The key is that direction matters. You can't just multiply A times B unless the field is perpendicular to the loop. (Check your text.)

Then dΦ/dt would be the derivative of the change in field/change in the time...

Right.

I didn't think the field was changing?

You are given that the field is a function of time. As t changes, B changes.

 

[dominatesean]

The key is that direction matters. You can't just multiply A times B unless the field is perpendicular to the loop. (Check your text.)

My text is of no help. All it does is confuse me even more. So I understand that if it's perp. then you don't need to use the cos of the dot product. I have no clue if it's perp. or not, all I get out of it is the box is an x/y cord. and the field is x/z. I used the dot product of those 2 (.11,.11,0)*(.31,0,.55) which gives me .0341 then I divide by change in time=.5, which I know is not right.

Thanks for your help by the way.

 

[Doc Al]

So I understand that if it's perp. then you don't need to use the cos of the dot product. I have no clue if it's perp. or not, all I get out of it is the box is an x/y cord. and the field is x/z.

You know the loop is in the x-y plane. So what coordinate is perpendicular to that?

I used the dot product of those 2 (.11,.11,0)*(.31,0,.55) which gives me .0341 then I divide by change in time=.5, which I know is not right.

(1) The area vector points normal to the loop surface, so it's not (.11,.11,0). Again, what direction is perpendicular to the loop (the x-y plane)?
(2) The magnetic field vector is not (.31,0,.55), it's (.31t,0,.55t²). (Since you need to know how that field changes, don't evaluate it at just one point. You'll evaluate the derivative at one point in time, since that tells you the rate of change.)
(3) Don't divide by the change in time, take the derivative.

 

[dominatesean]

I made that way too difficult and over thought everything.

Just the area * z
.0121*.55t²
take derivative and plug in for t

I'd like to thank Doc for his help & patience