Rate of thermal energy

[jena]

Hi,

My Question:

A 22.0 cm- diameter coil consists of 20 turns of circular copper wire 2.6 mm in diameter. A uniform magnetic field, perpendicular to the plane of the coil, changes at a rate of 8.65 x 10^-3 T/S.

Determine the rate of which thermal energy is produced.

My work:

P= i^2 x R
P= (.0013 A)^2(6.96 x 10^-4 ohms)
P= 1.18 x 10^-19 W

After this I'm lost. Should I look into trying to incorporate the rate, so that I can find the rate of which thermal energy is produced

Thank You:smile:

[Tide]

Where did your numbers for current and resistance come from?

[jena]

Got resistance by using

R=rho(L)/(A), where rho=1.68 x 10^-8 ohm*m, L= 22 x 10^-2 m, and A= pi((2.6 x 10^-3 m)/2)^2

so R= 6.96 x 10^-4 ohms

And to find the current I first had to find the Emf which I used the following equation below

Emf=-N(delta BA)/(delta t), where N=20 turns, and for the BA/t combo I used (-8.65 x 10^-3 T/s)(pi((2.6 x 10^-3 m)/2)^2)

Emf=-(20 turns)((-8.65 x 10^-3 T/s)(pi((2.6 x 10^-3 m)/2)^2)
)
Emf=9.19 x 10^-7 volts


Finally I used emf and R that I found to find the current

I=(emf)/R
I=(9.19 x 10^-7 volts)/(6.96 x 10^-4 ohms)
I=.0013 A

[Tide]

I didn't check your numbers but the argument looks good. The power you calculate IS the rate at which theremal energy is produced.

[lightgrav]

Jena,

the L in the resistance is the total length of the wire;
in your case, 20 turns x 2 pi (.11m) total length.
(so your R is too small by a factor 20 pi )

The changing B-field is encircled by an Electric Field,
where E 2 pi r = Delta(BA)/Delta(t) . Here,
A is the Area that is pierced by the changing B-field,
or the Area inside the encicling E-field loop (if smaller).

If E is parallel (along the LENGTH of) your COIL of wire,
the Voltage "accumulates" all along the wire, like
Delta(V) = E Delta(s) = N 2 pi R_coil .
So in this Delta V = N Delta(BA)/Delta(t) ,
the Area extends outward to the COIL of wire
(the place where the E-field makes a Voltage).
(If the coil is bigger than the B-field region,
(you only use the A where the B is going thru.)

Looks like you used the cross-section Area of the wire,
so your Area is too small by a factor of almost 10000.

[jena]

So to get the resistance I must first find the lenght

L=(N)(2 pi(22*10^-2 m/2)) and use this in the equation to help me find my resistance.

Is that what I'm supposed to do first.