Question

A circular loop of wire with a radius 9.712 cm is placed in a magnetic field such that it induces an emf of 3.18 v in the cir cular wire loop. if the cross-sectional diame ter of the wire is 0.333 mm, and the wire is made of a material which has a resistivity of 1.5×10−6 ▦ m, how much power is dissipated in the wire loop? answer in units of w.

Answer 1

To calculate the power dissipated in the wire loop, you need to find the current flowing through the wire and the voltage across the wire. Use Ohm's Law to calculate the resistance of the wire and then calculate the current using the induced EMF and resistance. Finally, multiply the current and voltage to find the power dissipated in the wire loop.

Step-by-step explanation:

First, let's calculate the current in the loop. We can use Ohm's Law, which states that V = IR, where R is resistance. The resistance of the wire loop can be calculated using the formula R = ρl/A, where ρ is resistivity, l is length, and A is the cross-sectional area of the wire. The length of the wire can be found using the formula l = 2πr, where r is the radius of the loop. Once we have the current and voltage, we can calculate the power dissipated in the wire loop by multiplying them together.

Let's plug in the values:

Radius of the loop (r) = 9.712 cm = 0.09712 m

Cross-sectional diameter of the wire = 0.333 mm = 0.000333 m

Resistivity (ρ) = 1.5 x 10^-6 Ωm

EMF induced (V) = 3.18 V

First, let's calculate the length of the wire:

l = 2πr = 2 x 3.14159 x 0.09712 = 0.610 m

Next, let's calculate the cross-sectional area of the wire:

A = πr^2 = 3.14159 x (0.000333/2)^2 = 2.1768 x 10^-7 m^2

Now, let's calculate the resistance of the wire:

R = ρl/A = (1.5 x 10^-6) x 0.610 / (2.1768 x 10^-7) = 4.1905 Ω

Finally, let's calculate the current:

I = V/R = 3.18 / 4.1905 = 0.7598 A

Now, we can calculate the power:

P = IV = 0.7598 x 3.18 = 2.418 W

Therefore, the power dissipated in the wire loop is 2.418 W.