Q. A horizontal wire $0.1 \,m$ long carries a current of $5 \,A$. The magnitude of the magnetic field which can support the weight of this wire is (assume mass of the wire as $3 \times 10^{-4}\, kg$ )

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A

$0.6 \times 10^{-3} T$

B

$10 \times 10^{-3} T$

C

$5.88 \times 10^{-3} T$

D

$2 \times 10^{-3} T$

Solution:

Force acting on acurrent carrying conductor due to magnetic field $=BIL$
Weight of the wire $=w=m g$
To support $F=W$
BIL$=mg B=m g / I L$
$B=3 \times 10^{-4} \times 9.8 / 5 \times 1$
$B=0.6 \times 10^{-3} T$

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