A straight wire is aligned north-south in a region where Earth\'s magnetic field

Question

A straight wire is aligned north-south in a region where Earth's magnetic field B directed 55.0 degree above the horizontal, with the horizontal component directed due north. The wire carries a current of 8.10 A toward the south. The magnetic force on the wire per unit length of wire has magnitude 260 times 10^-3 N_2/m. (a) What is the direction of the magnetic force on the wire? north south east west (b) What is the magnitude of B vector? A straight wire in a magnitude filed experiences a force of 0.28 N when the current in the wire is changed, and the wire experiences a force of 0.049 N as a result, What is the new current? A

Explanation / Answer

4 ans

Given that

angle=55 degree

current i=8.1 A

force/length (F/L)=2.6*10^-3 N/m

basing on the concept of the fleming left hand rule

now we find the magnetic field

Force F=BiLsin(theta)

F/L=Bi*sin(55)

2.6*10^-3=B*8.1*sin(55)

magnetic field B=3.92*10^-4 T

the magnetic force is north direction

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5 ans

Given that

length L=45 m

current i=64 A

force F=0.15 N

angle =62 degree

now we find the magnetic field

F=BiL sin(theta)

0.15=B*64*45*sin(62)

magnetic field B=5.9*10^-5 T

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6 ans

Given that

force F1=0.025 N

current i1=2.8 A

current i2=?

force F2=0.049 N

now we find the current i2

F1/F2=I1/I2

0.025/0.049=2.8/i2

current i2=5.5 A

therefore the current i2=5.5 A

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