Alength of 20-gauge copper wire (of diameter 0.8118 mm) is formed into a circular loop with a radius of 30.0 cm. a magnetic field perpendicular to the plane of the loop increases from zero to 14.0 mt in 0.20 s. find the average electrical power dissipated in the process.(the resistivity of copper is 1.72 × 10 − 8 1.72×10−8 ω . m ω.m)

3 x 10^-24 watt

Explanation:

P ( resistivity) = 1.72e-8 (from the chart).

L= 2pi r

r= 30 cm.

R= pL/A

A= pi* r1^2

r1= 0.8118/2 * 10^-3 m

R= 1.68 x 10^-8 x (2x3.142x0.3)

= 3.24 x 10^-8

E=N do/dt

do= B* A

A= pi* 0.3^2

N=1

E = 1 x (14 x 3.142x 0.09) = 3.95

I=v/R

v=E,

I = 3.95 / 3.24 x 10^-8 = 1.22 x 10^8

P=I^2 x R.

= 3 x 10^-24 watt

a solenoid is created by wrapping a l = 90 m long wire around a hollow tube of diameter d = 4.5 cm.   the wire diameter is d = 0.9 mm.   the solenoid wire is then connected to a power supply so that a current of i = 9 a flows through the wire.

randomized variablesl = 90 m

d = 4.5 cm

d = 0.9 mm

i = 9 a                                          

write an expression for the number of turns, n, in the solenoid. you do not need to take into account the diameter of the wire in this calculation.

calculate the number of turns, n, in the solenoid.

write an expression for the length of the solenoid (l2) in terms of the diameter of the hollow tube d.

calculate the length of the solenoid (l2)   in meters.

calculate the magnitude of the magnetic field at the center of the solenoid in teslas.

explanation: