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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.

Sagot :

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  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 sinθ

Where;

  • θ 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 10⁻²) x (48) x 24 x sin(90)

F = 41.472 N

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

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

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