A wire carrying current $I$ and length $l$ is placed in a uniform magnetic field of magnitude $B$ in such a manner that the wire is suspended in air due to the magnetic force.
The mass of the wire is:

B I l g

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B l g

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\frac{g}{B I l}

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\frac{B I l}{g}

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B I l

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Solution

The correct option is D $\frac{B I l}{g}$
Here the weight $(m g)$ of the object is balance by the magnetic force $F_{B} = B I L$
Thus, $m g = B I L$
or $m = B I L / g$
Thus, option D will correct.

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A wire carrying current I and length l is placed in a uniform magnetic field of magnitude B in such a manner that the wire is suspended in air due to the magnetic force.The mass of the wire is:

The correct option is D BIlgHere the weight (mg) of the object is balance by the magnetic force FB=BILThus, mg=BILor m=BIL/gThus, option D will correct.

A wire carrying current $I$ and length $l$ is placed in a uniform magnetic field of magnitude $B$ in such a manner that the wire is suspended in air due to the magnetic force.
The mass of the wire is:

The correct option is D $\frac{B I l}{g}$
Here the weight $(m g)$ of the object is balance by the magnetic force $F_{B} = B I L$
Thus, $m g = B I L$
or $m = B I L / g$
Thus, option D will correct.