Physics magnetics?

Question

A piece of copper wire is formed into a single circular loop of radius 9 cm. A magnetic field is oriented parallel to the normal to the loop, and it increases from 0 to 0.45 T in a time of 0.45 s. The wire has a resistance per unit length of 3.3 10-2 /m. What is the average electrical energy dissipated in the resistance of the wire.

Answer 1

A changing B-field produces “induced emf” in a coil placed therein. A circular loop of radius (r= 0.09 m) when placed in a magnetic field B experiences a magnetic flux depending upon the relative orientation of B and area vector of coil.
Flux Phi = B*A cos (p) when p=0 then Phi = B*A

where B (weber = T- m^2) and (p) is angle between B and area vector. When B is parallel to area vector then p=0. NOW the flux can change either by rotating the coil (dA - through p angle) or by changing field (dB). Here, the latter case is applicable because (p=0) and coil is fixed.

As per Faraday’ law, the changing flux (in given time t) through the coil (of N turns) produces an Induced emf which is equal to
(emf)average = - N [ Delta (phi) ] / Delta (t) ------(1)

(emf)rms = - N [ d (phi) ] / dt ------(2)

The emf given by (1) is average emf (to be found) and negative is just to indicate that it is back emf (which tries to maintain the flux). But when B is varying with time in sine wave then differentials are used, that give “rms” (root mean square) values by which (emf)max can be calculated from (2).

N = 1, Area = Pie*r^2=3.14*(0.09)^2 = 0.02540 m^2,

(emf)av = - A* [B (2)–B (1)] / Delta (t) = (0.0254* 0.45) / 0.45
= 0.0254 volts = 25.4 milli volts

The Resistance of copper wire coil
R = rho * L = 3.3* 10^-2 (ohm/m) * 2 Pi * 0.09 = 0.01865 ohms

The average electrical energy dissipated due the resistance of the coil countering induced current in given time =
(Energy)av = (Power)av * delta (t) = [{emf)av}^2 *delta (t)] / R
= (0.0254)^2 * 0.45 /0.01865 = 15.56 *10^-3 joule

Answer 2

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