contestada

What is the size of the force acting on a copper wire with a magnetic flux density of 3.6 x 10-2 T acting at
right angles to the wire of length 24 m and current of C:048 A? Give your answer to an appropriate
number of significant figures.

Respuesta :

Answer:

F = 0.414 N

Explanation:

Given that,

Magnetic flux density,[tex]B=3.6\times 10^{-2}\ T[/tex]

The length of the wire, l = 24 m

Current, I = 0.48 A

We need to find the force acting on the wire. The formula for the force is given by:

[tex]F=ILB[/tex]

Put all the values,

[tex]F=0.48\times 24\times 3.6\times 10^{-2}\\\\F=0.414\ N[/tex]

So, the force acting on the copper wire is equal to 0.414 N.

onyebuchinnaji

The magnetic force of the copper wire is Ā 41.472 N.

Magnetic force of the copper wire

The magnetic force of the copper wire is calculated by applying the following equation.

F = BIL x sinĪø

Where;

  • Īø is the inclination of the magnetic field
  • I is the current
  • L is the length of the wire
  • B is the magnetic field strength = flux density

F = (3.6 x 10ā»Ā²) x (48) x 24 x sin(90)

F = 41.472 N

Thus, the magnetic force of the copper wire is Ā 41.472 N.

Learn more about magnetic force here: Ā https://brainly.com/question/13277365

Otras preguntas