A wire in the shape of an \"M\" lies in the plane of the paper. It carries a cur

Question

A wire in the shape of an "M" lies in the plane of the paper. It carries a current of 2.0 A, flowing from points A to E, as shown in the figure. It is placed in a uniform magnetic field of 0.75 T in the same plane, directed as shown on the right side of the figure. The figure indicates the dimensions of the wire.

(a) What is the magnitude of the force acting on section AB of this wire?

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Part B

(b) What is the direction of the force acting on section AB of this wire?

(b) What is the direction of the force acting on section AB of this wire?

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Part C

(c) What is the magnitude of the force acting on section BC of this wire?

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Part D

(d) What is the direction of the force acting on section BC of this wire?

(d) What is the direction of the force acting on section BC of this wire?

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Part E

(e) What is the magnitude of the force acting on section CD of this wire?

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Part F

(f) What is the direction of the force acting on section CD of this wire?

(f) What is the direction of the force acting on section CD of this wire?

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Part G

(g) What is the magnitude of the force acting on section DE of this wire?

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Part H

(h) What is the direction of the force acting on section DE of this wire?

(h) What is the direction of the force acting on section DE of this wire?

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Part I

(i) What is the magnitude of the force acting on the entire wire?

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Part J

(j) What is the direction of the force acting on the entire wire?

(j) What is the direction of the force acting on the entire wire?

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Explanation / Answer

Fb = I*L*B*sintheta

part A)

a) Fb = 2*0.1*0.75*sin45 = 0.106 N     <<<----------answer

part B)

b) perpendicular out of the page        <<<----------answer

part C)

c) Fb = 2*0.1*0.75*sin0 = 0       <<<----------answer

part D)

d) none of the above    <<<----------answer

part E)

Fb = 2*0.1*0.75*sin90 = 0.15 N <<<----------answer

part F)

F) perpendicular out of the page

part G)

g) FB = 2*0.1*0.75*sin45 = 0.106 N

part H

h) perpendicular into the page

part I

i) FCD = Fb = 2*0.1*0.75*sin90 = 0.15 N

part J

j) perpendicular out of the page

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