Respuesta :
Answer:
[tex]1.27\times 10^{-4}N[/tex]
Explanation:
We have given
Length of wire  l=20 mm=[tex]20\times 10^{-3}m[/tex]
Current carried by solenoid i =3 A
Number of turns n = 600 turns per meter
Magnetic field inside a solenoid is given by [tex]B=\mu _0ni=4\pi \times 10^{-7}\times 600\times 3=0.0022608T[/tex]
Current carried by the wire = 4 A
The wire is suspended at an angle of 45° so [tex]\Theta =45^{\circ}[/tex]
Now force on the wire [tex]F=IBLSIN\Theta =4\times 0.002260\times 20\times 10^{-3}\times SIN45^{\circ}=1.27\times 10^{-4}N[/tex]
The magnitude of the magnetic force exerted on a wire when it is suspended inside the solenoid is [tex]1.28\times10^{-4}\rm N[/tex].
What is magnetic force?
Magnetic force is the force of attraction of repulsion between two poles of the two magnets. The magnetic force is also appears between two electrically charged bodies.
The magnetic force can be given as,
[tex]F=IBL\sin\theta[/tex]
Here, (I) is the current carried by wire , (B) is the strength of the magnetic field, (L) is the length of the wire and ([tex]\theta[/tex]) is the angle between the magnetic field and wire.
The magnetic filed inside a solenoid can be given as,
[tex]B=4\pi\times10^{-7}nI[/tex]
Here, (n) is the number of turns in a solenoid and (I) is the current carried by it.
As the total number of turns the solenoid has is 600 per meter of length and the current carried by the solenoid is 3.0 A. Thus the magnetic filed inside the solenoid can be given as,
[tex]B=4\pi\times10^{-7}\times600\times3\\B=2.2608\times10^{-3}\rm T[/tex]
Now the wire with 20 mm is suspended inside the solenoid at an angle of 45∘. The current carried by the wire is 4.0 A. Thus the magnetic force on the wire can be given as,
[tex]F=4\times2.2608\times10^{-3}\times20\times10^{-3}\times\sin(45)\\F=1.28\times10^{-4}\rm N[/tex]
Thus the magnitude of the magnetic force exerted on a wire when it is suspended inside the solenoid is [tex]1.28\times10^{-4}\rm N[/tex]
learn more about the magnetic force here;
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