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A loop of wire carrying a current of 2.0 A is in the shape of a right triangle with two equal sides, each 15 cm long. A 0.7 T uniform magnetic field is parallel to the hypotenuse. Find the magnitude of the resultant magnetic force on the two sides.

Respuesta :

Answer:

F =  0.312 N

Explanation:

Given,

Current, I = 2 A

Length of the equal side = 15 cm = 0.15 m

Magnetic field, B = 0.7 T

Magnetic filed is parallel to hypotenuse

θ = 135°

Force on the first side of the triangle

[tex]F = i BL \sin \theta[/tex]

Force on the another side

[tex]F = i BL \sin \theta[/tex]

Resultant magnetic Force

[tex]F = 2 \times 2\times 0.7 \times 0.15 \sin 135^0[/tex]

F =  0.312 N

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