4/18/2018 11:59 PM 10/100 Gradebo PrintCalulator Periodic Table uestion 7 of 10

Question

4/18/2018 11:59 PM 10/100 Gradebo PrintCalulator Periodic Table uestion 7 of 10 A wire of length 0.35 m is a current of I-75A Map in the +x direction through a B- 3.0-T uniform magnetic field that is directed parallel to the wire in the-x direction. What are the magritude and direction of the magnetic force on the wire? Number #? -y O none of the above + x field changes, so that now the force on the wire is F4.0 N in the *y direction. What are the magnitude and direction of the field now? (Assume that the magnetic field B is perpendicular to L.) Number +2 a none of the Next -HE -O Previous 9Check Answer

Explanation / Answer

F = IL x B

F = I L B sin(theta)

theta = 180 deg and sin180 = 0

so F =0 N ......>Ans

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F = 4 j^

and IL = (0.35 x 7.5) i^

F = IL x B

4 j^ = 2.625 i^ x B

B = 1.52 T (-k^)

Ans: 1.52 T And - z  

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