Horizontal wire of. 1 m long carries a current of 5 ampere find the magnitude and direction of the magnetic field which can support the weight of the why they went the mass of the wire is 3 into 10 power minus 3 kg per metre
Answers
Answered by
4
Let B be uniform magnetic field ,current in the wire of length L be I, then force due to magnetic field is given by
F=ILxB.
L=0.5m
I=1.2 A
B=2T
Therefore,
|F|=(1.2)(0.5)(2)(sin pi/2)=1.2N. In the direction perpendicular to the plane formed by LandB according to right hand screw law for cross product of two vectors.
F=ILxB.
L=0.5m
I=1.2 A
B=2T
Therefore,
|F|=(1.2)(0.5)(2)(sin pi/2)=1.2N. In the direction perpendicular to the plane formed by LandB according to right hand screw law for cross product of two vectors.
Answered by
10
Answer:
In equilibrium
magnetic force on wire= weight of wire
or, IlB sin 90°= mg
B=mg/I l =3x 10^-3 x 9.8/5
= 5.88 x10^-3 T
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