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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.
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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