Question

A straight vertical wire carries a current of 1.60 AA downward in a region between the poles of a large electromagnet where the field strength is 0.550 TT and is horizontal.

Part A

Part complete

What are the direction of the magnetic force on a 1.20 cmcm section of this wire if the magnetic-field direction is toward the east?

The direction is southward.

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Correct

Part B

What are the magnitude of the magnetic force on a 1.20 cmcm section of this wire if the magnetic-field direction is toward the east?

Express your answer in newtons.

FF =

0.153

NN

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Part C

Part complete

What are the direction of the magnetic force on a 1.20 cmcm section of this wire if the magnetic-field direction is toward the south?

The direction is westward.

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Correct

Part D

What are the magnitude of the magnetic force on a 1.20 cmcm section of this wire if the magnetic-field direction is toward the south?

Express your answer in newtons.

FF =

nothing

NN

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Part E

What are the direction of the magnetic force on a 1.20 cmcm section of this wire if the magnetic-field direction is 30.0 oo south of west ?

Express your answer as an angle in degrees north of west.

ϕϕ =

nothing

∘∘

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Part F

What are the magnitude of the magnetic force on a 1.20 cmcm section of this wire if the magnetic-field direction is 30.0 oo south of west ?

Express your answer in newtons.

FF =

nothing

NN

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Answer 1

given

current of 1.60 A = i

field strength is 0.550 T and is horizontal = B

r = 1.2 cm

= 0.012 m

Part A )

according to Right Hand Rule,

the direction of the current is in down and magnetic field direction is in east and the direction is in

SOUTH direction.

Part B )

F = B i L sin

= 0.55 x 1.6 x 0.012 x sin90

F = 0.01056 N

Part C )

the current direction is in DOWN and magnetic field direction is in SOUTH, force direction is WEST.

Part D )

F = B i L sin90

= 0.55 x 1.6 x 0.012 x sin90

F = 0.01056 N

Part E )

current direction is DOWN, magnetic field direction 30^o SOUTH of WEST, force direction is 60^o

NORTH of WEST.

Part F )

F = B i L sin90

= 0.55 x 1.6 x 0.012 x sin90

F = 0.01056 N