A straight horizontal conductor of length $L$ meters and mass $m$ kg carries a current $i$ ampere. The minimum magnetic induction which must exist in the region to balance its weight is :
(Given Acceleration due to gravity is $g$ )

\frac{m g}{i L}

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\frac{i L}{m g}

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\frac{m g L}{i}

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\frac{m L}{i g}

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Solution

The correct option is B $\frac{m g}{i L}$
Force acting on a current carrying wire $= i L B$
Force due to weight $= m g$
So, to balance weight
$F$ . due to current $= F$ due to weight
$\Rightarrow i L B = m g$
$\Rightarrow B = \frac{m g}{i L}$

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A straight horizontal conductor of length L meters and mass m kg carries a current i ampere. The minimum magnetic induction which must exist in the region to balance its weight is : (Given Acceleration due to gravity is g)

The correct option is B mgiLForce acting on a current carrying wire =iLBForce due to weight =mgSo, to balance weight F. due to current =F due to weight⇒iLB=mg⇒B=mgiL

A straight horizontal conductor of length $L$ meters and mass $m$ kg carries a current $i$ ampere. The minimum magnetic induction which must exist in the region to balance its weight is :
(Given Acceleration due to gravity is $g$ )

The correct option is B $\frac{m g}{i L}$
Force acting on a current carrying wire $= i L B$
Force due to weight $= m g$
So, to balance weight
$F$ . due to current $= F$ due to weight
$\Rightarrow i L B = m g$
$\Rightarrow B = \frac{m g}{i L}$