### Simple Metal Detector Using Coils**Diagram Description:**- The diagram consists of two circular coils of wire.- **Inner Coil**: - Labeled "Inner coil, radius R2 with N2 turns". - Used to measure voltage. - Significantly smaller than the outer coil. - **Outer Coil**: - Labeled "Outer coil, radius R1 with N1 turns, ΔI/Δt = 5 amps/second". - Current increases at a rate of 5 amps per second.**Concept Explanation:**We construct a simple metal detector using two circular wire coils. The inner coil, which has a radius of R2 and N2 number of turns, is much smaller than the outer coil (with radius R1 and N1 turns). This size difference allows us to assume the magnetic field remains constant over the inner coil.### QuestionWhen the outer coil is driven with an increasing current (5 amps/second), and there is no metal present, how much voltage is induced in the inner coil and why?### Understanding the Induced VoltageThe voltage induced in the inner coil can be determined using Faraday's Law of electromagnetic induction, which states:\[\text{Induced Voltage} (\epsilon) = -N \frac{\Delta \Phi}{\Delta t}\]- \(N\) is the number of turns in the coil.- \(\Delta \Phi\) is the change in magnetic flux.- \(\Delta t\) is the change in time.Given that the rate of change of current (ΔI/Δt) is 5 amps/second in the outer coil, this change in magnetic field induces a voltage in the smaller inner coil. The voltage measured depends on the number of turns and the rate of change of the magnetic field passing through it.This simulation helps in understanding the basic principle of metal detectors and electromagnetic induction.

The field due to outer coil in place of inner coil is, B=μ04π2πIR1N1 Since the flux through inner…

Inner coil, radius R2 with N2 turns measure voltage Outer coil, radius R1 with N1 turns, AI/At = 5 amps/second We make a simple metal detector out of two circular coils of wire. The inner coil (radius R2, number of turns N2) is much much smaller than the outer coil (radius R1, number of turns N1), so you can assume the magnetic field is constant over the inner coil. We drive the outer coil with a current that increases at 5 amps/second. If there is no metal around, how much voltage is induced on the inner coil and how do you know? 1.

Inner coil, radius R2 with N2 turns measure voltage Outer coil, radius R1 with N1 turns, AI/At = 5 amps/second We make a simple metal detector out of two circular coils of wire. The inner coil (radius R2, number of turns N2) is much much smaller than the outer coil (radius R1, number of turns N1), so you can assume the magnetic field is constant over the inner coil. We drive the outer coil with a current that increases at 5 amps/second. If there is no metal around, how much voltage is induced on the inner coil and how do you know? 1.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Similar questions

Chapter 30, Concept Question 04 The figure shows two circuits in which a conducting bar is slid at the same speed v through the same uniform magnetic field and along a U- shaped wire. The parallel lengths of the wire are separated by 24 in circuit 1 and by L in circuit 2. The current induced in circuit 1 is counterclockwise. e (2) Is the magnetic field into or out of the page? Into Out of Is the current induced in circuit 2 clockwise or counterclockwise? Counterclockwise Clockwise Is the emf induced in circuit 1 larger than, smaller than, or the same as that in circuit 2? Larger The same as Smaller
QUESTION 1 A coil of ten turns of area 0.01 m2 and resistance 5 Q is placed perpendicularly inside a magnetic field whose intensity and direction changes as given in the figure. What is the magnitude of the flux of magnetic field crossing the area of the coil at time = 6 sec? B(T) 0.1 0.3 0.5 0.8 0.9 t(s) -1 -2 O 2 Wb O 0.2 Wb O 20 Wb O 0.02 Wb O none of these
The magnetic field at the center of a solenoid L = 75 cm long is B = 0.11 T when a current of I = 3.5 A flows through the solenoid wire. 1. Solve the formula for the magnetic field near the center of a long, tightly wound solenoid for the number of turns, N, in the solenoid. The expression should be in terms of the given variables. 2. Calculate the value of the number of turns, N.
A stationary square coil of area 0.1 m² is brought into the magnetic field 5.5 T with its plane perpendicular to the magnetic field. The coil has 1363 turns. Calculate the magnetic flux linkage through the coil. Give your answer in Sl units. Answer: Choose...
A small circular coil of 20 loops of wire having a radius of 1 cm is at rest in the X-y plane (so the normal to the loop points in the +z-direction). A uniform magnetic field of 0.5 T is turned on in the +x-direction. What is the magnetic flux through the loop? a. 7.85 x 10 Tm² b. 1.57 x 10 4 T m² O c. 3.14 x 10 3 T m2 O d. 0 T m2
23 and 24 please
I need the answer as soon as possible
I need the answer as soon as possible
G7
Need Help? mV Sascha Hahn/Shutterstock.com A helicopter (see figure below) has blades of length 3.40 m, extending out from a central hub and rotating at 2.30 rev/s. If the vertical component of the Earth's magnetic field is 50.0 μT, what is the magnitude of emf induced between the blade tip and the center hub? Read It
I need the answer as soon as possible
Q. What are the values of (a) L and 5. The secondary coil of a step-up transformer provides the voltage that operates an electrostatic air filter. The turns ratio of the transformer is 50:1. The primary coil is plugged into a standard 120-V outlet. The current in the secondary coil is 1.7 x 103 A. Find the power consumed by the air filter. 6. The earth's magnetic field, like any magnetic field, stores energy. The maximum strength of the