# Inductive Heater Homework: Find Current & Rate of Energy

• Angie K.
In summary: The total length of wire is 25*2*pi*.173 = 21.624 meters.So R = p (L/A) = 1.68*10^-8 (21.624 / pi*(0.00225/2)^2 = 1.920157Then I = Emf/R = B*Area of coil / R = 3.41194*10^-2 / 1.920157 = 17.774 AIn summary, a 34.6-cm-diameter coil with 25 turns of circular copper wire with a diameter of 2.25 mm is placed in a uniform magnetic field changing at a rate of 3.55
Angie K.

## Homework Statement

A 34.6 -cm-diameter coil consists of 25 turns of circular copper wire 2.25 mm in diameter. A uniform magnetic field, perpendicular to the plane of the coil, changes at a rate of 3.55 × 10-3T/s. Determine

a. the current I in the loop,

b. and the rate dU/dt at which thermal energy is produced.

## Homework Equations

magnetic flux = B*A cos Theta

I = magnetic flux / R

R = p (L/A) where p is the resistivity of copper

## The Attempt at a Solution

Here is what I attempted:

First, I found the magnetic flux by using the formula: B*A cos Theta

multiplying the area by 25 because it's in a coil and using .173 meters for the radius of the coil

(3.55*10^-3)(25*.173) = 1.535375*10^-2

then to find I (current) :

I = magnetic flux / resistance

I calculated resistance by :

since it's a coil, I multiplied the L by 25 and the A by 25

R = p (L/A) = 1.68*10^-8 (25*pi*.173 / 25*pi*1.125*10^-2)^2 = 0.001148

plugging in that into the I equation, I = 1.535375e-2 / 0.001148 = 13.374 A

This isn't the right answer and I messed something up. I would appreciate if someone could help me with catching my mistake! I appreciate it!

again. why do you say that i=magnetic flux/R ?
##E=-d\phi/dt##.
so i is the change in magnetic flux per unit time/R.
what is the area of cross section of wire?

again. why do you say that i=magnetic flux/R ?
##E=-d\phi/dt##.
so i is the change in magnetic flux per unit time/R.
what is the area of cross section of wire?

I meant to say that i = emf/R

The area of the cross section is pi*r^2 which is pi(.00225)^2 where .00225 is the radius of the wire

= 1.590431e-5

Would I only multiply the circumference of the coil by 25?

no. 2.25 is its diameter.

no. 2.25 is its diameter.

R = p (L/A) = 1.68*10^-8 (25*pi*.173 / pi*(0.00225/2)^2 = .11482

then I = Emf/R = B*Area of coil / R = 3.41194*10^-2 / .11482 = 0.297155 A

Still not getting it right. When calculating for the Emf, did I do that right? or is the fact that i's a changing magnetic field need some different calculation to get the Emf?

what is the total length of wire? your value of L in R=p(LA) is wrong.

## 1. How does an inductive heater work?

An inductive heater works by using electromagnetic induction to generate heat. It consists of a high-frequency alternating current (AC) that passes through a coil of wire, creating an oscillating magnetic field. This magnetic field then induces eddy currents in a conductive material, such as a metal pan, causing it to heat up.

## 2. What is the difference between current and rate of energy in an inductive heater?

Current refers to the flow of electricity in an inductive heater, measured in amperes (A). Rate of energy, on the other hand, refers to the amount of energy being transferred per unit of time, usually measured in watts (W). In other words, current is the cause of energy transfer, while rate of energy is the effect.

## 3. How do you find the current in an inductive heater?

To find the current in an inductive heater, you can use Ohm's law, which states that current is equal to voltage divided by resistance (I = V/R). In an inductive heater, the voltage is typically known and the resistance can be calculated based on the material and dimensions of the coil and the conductive material being heated.

## 4. How do you calculate the rate of energy in an inductive heater?

The rate of energy in an inductive heater can be calculated using the formula P = I^2 * R, where P is power (in watts), I is current (in amperes), and R is resistance (in ohms). This formula takes into account the relationship between current and resistance in generating heat.

## 5. What factors can affect the current and rate of energy in an inductive heater?

The current and rate of energy in an inductive heater can be affected by various factors, such as the strength and frequency of the alternating current, the size and shape of the coil, the type and size of the conductive material, and any external factors that may interfere with the magnetic field, such as nearby metal objects or electromagnetic radiation.

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