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A current-carrying wire of length 52.0 cm is positioned perpendicular to a uniform magnetic field. If the current is 15.0 A and it is determined that there is a resultant force of 2.3 N on the wire due to the interaction of the current and field, what is the magnetic field strength?

Respuesta :

Answer:

The magnetic field strength is 0.29 T.

Explanation:

Given that,

Length of current-carrying wire, L = 52 cm = 0.52 m

Current, I = 15 A

Magnetic force, F = 2.3 N

We need to find the magnetic field strength. We know that the magnetic force is given by :

[tex]F=ILB[/tex]

B is magnetic field strength

[tex]B=\dfrac{F}{IL}\\\\B=\dfrac{2.3}{0.52\times 15}\\\\B=0.29\ T[/tex]

So, the magnetic field strength is 0.29 T.

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