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Q. A wire of mass 100gm, length 1m and current 5 ampere is balanced in mid air by a magnetic field B, then find the value of B.

Naina Singh , 8 Years ago
Grade 12th pass
anser 2 Answers
Khimraj

Last Activity: 8 Years ago

ILB = mg5*1*B =(1/10)10B =1/5 T. Hope it clears. If u have doubt then please ask.Please approve the answer.

Shreya D Nanda

Last Activity: 8 Years ago

Hi,
As the wire is balanced in mid air, the downward force is equal to the upward force, i.e, the gravitational force is equal to the magnetic force due to the magnetic field B.
Mg=BIL \sin \theta
as B is perpendicular to the current carrying wire, \theta = 90 \Rightarrow sin\theta =1
\therefore Mg= BIL
sustituting the values, in SI units, { g = 9.8m/s2 }
0.1 * 9.8 = B * 5 * 1
B= 0.98/5 T
or B= 0.196 T.
 
hope it is understandable,...
 

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