A horizontal wire of length 1.0 m carries a current of 48 A to use right. A seco

Question

A horizontal wire of length 1.0 m carries a current of 48 A to use right. A second wire with a length of 0.5 m and a mass of 0.50 kg is suspended 0.30 m above the first wire. A.) In what direction (left or right) does to current in the upper wire have to flow in order to be suspended agents the of gravity? B) What is the magnitude of the magnetic field (in mu T) due to the bottom wire at the location of the top wire? C) What is the weight (in N) of the top wire that must be balanced by the magnetic field? D) What is the magnitude of the current in the top wire (in mega Amps)?

Explanation / Answer

A)Since the upper wire will experience magnetic force due to currrent in below wire .The weight of upper wire will act downwards ,so magnetic force must be upward to balance it .Since current in below wire is to the right ,magnetic field in upper wire will be out of the plane of paper .so current in above wire must be toward left .to make magnetic force upward

Direction=Left

B)Magnetic field=u*I/2*pi*r=3.2*10^(-5) T

C)Weight=mg=4.9 N

(D) let current be i

force=iLB=i*0.5*B=4.9

i=306250 A=0.306 MA

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