Calculating Energy Dissipated in Copper Wire Loop Due to Magnetic Field Increase

[StudentofPhysics]

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.


I know the emf = 0.254 Volts.

How do I find the energy dissipated?

 

[Doc Al]

Start by finding the current that flows through the wire.

 

[kyle8921]

As a scientist, it is important to understand the relationship between magnetic fields and electrical currents. In this scenario, the increasing magnetic field induces a current in the copper wire loop, which in turn causes the wire to dissipate energy in the form of heat due to its resistance. To calculate the average electrical energy dissipated in the resistance of the wire, we can use the formula:

E = I^2Rt

Where E is the energy dissipated, I is the current, R is the resistance per unit length, and t is the time.

First, we need to calculate the current (I) in the wire using Ohm's law:

I = V/R

Where V is the emf (0.254 V) and R is the resistance per unit length (3.3 x 10^-2 Ω/m).

Substituting these values, we get I = 0.254 V / 3.3 x 10^-2 Ω/m = 7.7 A.

Next, we need to calculate the total resistance of the wire loop. Since it is a circular loop, we can use the formula for the resistance of a wire in a circular loop:

R = ρL/A

Where ρ is the resistivity of copper (1.68 x 10^-8 Ωm), L is the length of the wire (circumference of the loop = 2πr = 2π(0.09 m) = 0.565 m), and A is the cross-sectional area of the wire (πr^2 = π(0.09 m)^2 = 0.0257 m^2).

Substituting these values, we get R = (1.68 x 10^-8 Ωm)(0.565 m) / 0.0257 m^2 = 3.7 x 10^-6 Ω.

Now, we can plug in these values into the original formula to calculate the energy dissipated:

E = (7.7 A)^2 (3.7 x 10^-6 Ω) (0.45 s) = 1.5 x 10^-4 J

Therefore, the average electrical energy dissipated in the resistance of the wire is 1.5 x 10^-4 Joules. This energy is converted into heat, which can be measured by the temperature increase of the wire.