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Image for A wire carrying a current 30 A has a length 0.1 m between the pole faces of a magnet at an angle 60 Degree (se

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Answer #1

answer

F= Il ×B

F(304 (0.1m) (0.5T)sin 60199N -1.299N

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Answer #2

here

current I = 30A

length l =0.1 m

angle \Theta =60degrees

magnetic field B = 0.5T

F= ilB sin\Theta

plugging in the values we get

F =1.3N

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