Answered: The principle of induction heating is… | bartleby

Related questions

Question

The principle of induction heating is explored in this problem. A sheet of
metal (good conductor) of conductivity oc is inserted into AC magnetic field
whose value at the surface z =+d and
wwww
H(d) = (-unitvector y) 10 A/m
H(-d) = (-unitvector y) 20 A/m
Determine the expressions for (a) the magnetic field vector field H (z) and (b) the
volume current density field J (z) in the field region -d <z< d.
H10
d
-d

Expert Solution

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Blurred answer

Similar questions

An oscillating LC circuit consisting of a 1.3 nF capacitorand a 4.5 mH coil has a maximum voltage of 2.3 V.What are (a) themaximum charge on the capacitor, (b) the maximum currentthrough the circuit, and (c) the maximum energy stored in themagnetic field of the coil?
An MRI technician moves his hand from a region of very low magnetic field strength into an MRI scanner’s 2.00 T field with his fingers pointing in the direction of the field. Find the average emf induced in his wedding ring in mV, given its diameter is 2.2 cm and assuming it takes 0.45 s to move it into the field.
The 5-cm by 8-cm rectangular coil of a bicycle generator is rotating at a rate of 125 rpm in a 255-mT magnetic field. The output voltage of the generator is used to power a bicycle headlight that has a total resistance of 360 N. (a) How many turns are there in the coil if the peak output voltage is 1.8 V? N = turns (b) What is the rms-average current through the headlight? Irms mA
A solenoid with 500 turns and a cross-sectional area of 0.01 m² is placed in a magnetic fieldthat increases from 0 T to 1 T in 4 seconds. Calculate the average induced EMF in the solenoid.
An MRI technician moves his hand from a region of very low magnetic field strength into an MRI scanner's 2.10 T field with his fingers pointing in the direction of the field. His wedding ring has a diameter of 2.11 cm, and it takes 0.320 s to move it into the field. (a)What average current is induced in the ring if its resistance is 0.0100 Ω? (Enter the magnitude in amperes.) A (b)What average power is dissipated (in W)? W (c) What average magnetic field is induced at the center of the ring? (Enter the magnitude in teslas.)T
(d) Calculate the magnitude of torque (in Nm) exerted by the magnetic field on the coil at the instant when the emf is a maximum. N.m (e) What is the maximum power delivered to the coil (in W)? W (f) What If? If the coil is constructed from aluminum wire, what is the cross-sectional area of the wire (in m²)? The resistivity of aluminum is 2.82 x 10-8 Q. m. m² (g) What If? Suppose the coil now has the same number of turns, dimensions, and resistance, but the maximum current that the wire can now safely carry is 4.50 A. What now is the maximum angular speed (in rad/s) with which the coil can be rotated in the same 2.20 T magnetic field? rad/s
A 400 Hz alternating magnetic field produces a 0.5 V rms output from a 10-turn search coil with a diameter of 10 cm. Calculate the peak field flux density.
In the figure below, a semicircular conductor of radius R = 0.220 m is rotated about the axis AC at a constant rate of 170 rev/min. A uniform magnetic field of magnitude 1.32 T fills the entire region below the axis and is directed out of the page. R Воли (a) Calculate the maximum value of the emf induced between the ends of the conductor. (b) What is the value of the average induced emf for each complete rotation? (c) How would your answers to parts (a) and (b) change if the magnetic field were allowed to extend a distance R above the axis of rotation? (Select all that apply.) O The value in part (a) would increase. The value in part (a) would remain the same. The value in part (a) would decrease. The value in part (b) would increase. The value in part (b) would remain the same. The value in part (b) would decrease. (d) Sketch the emf versus time when the field is as drawn in the figure. Choose File no file selected (e) Sketch the emf versus time when the field is extended as…