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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.

Grade:12th pass

## 2 Answers

Khimraj
3007 Points
4 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
74 Points
4 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.
$\dpi{80} Mg=BIL \sin \theta$
as B is perpendicular to the current carrying wire, $\dpi{80} \theta = 90 \Rightarrow sin\theta =1$
$\dpi{80} \therefore Mg= BIL$
sustituting the values, in SI units, { g = 9.8m/s2 }
$\dpi{80} 0.1 * 9.8 = B * 5 * 1$
$\dpi{80} B= 0.98/5 T$
or $\dpi{80} B= 0.196$ T.

hope it is understandable,...

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