A 9.00-cm-long wire is pulled along a U-shaped conducting rail in a perpendicular magnetic field. The total resistance of the wire and rail is 0.320 Ω . Pulling the wire at a steady speed of 4.00 m/s causes 4.30 W of power to be dissipated in the circuit.

0.45 N

Explanation:

Let Recall that,

The power formula is:  

  P = E²/R  

Let A = the magnetic field  

Let L = length of wire  = 9.00cm = 0.09 m  

let  R = resistance of wire  = 0.320 Ω

let v =  velocity of the wire = 4 m/s  

Let E = across the wire voltage

Let P = the power of the wire = 4.3 W  

To Solve  for E:  

The formula of E = √PR  

The Voltage from  a magnetic field is given as,

E = vAL  

We therefore Use E = E

√PR = vAL  

to solve  for A,

A= √PR/vL  

BA= √4.3(0.32)/(4)(.09)  -=0.173

A = 0.173 wA/m²

Let F  be  the pulling force  

Let I  be the current in the wire  

P = I²R  

I = √P/R  

F = IAL  

F = √P/RAL  

F = √4.3/.32(0.173)(.09)  

F = 0.45N

when a device is used in a circuit in which the voltage is 81 v, the current flowing through the device is 3 a. what is the resistance of the device?

a. 27 ω

b. 40.5 a

c. 243 ω

d. 243 a

the correct answer is a. 27

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this