Question

A horizontal wire 0.1 m long carries a current of 5 A. Find the magnitude of the magnetic field, which can support the weight of the wire. Assume wire to be of mass $3\times {10}^{-3}kg{m}^{-1}$ :

• A. $5.88\times {10}^{-2}T$
• B. $4.88\times {10}^{-3}T$
• C. $5.88\times {10}^{-3}T$
• D. $5.88\times {10}^{-4}T$

Answer 1

The correct option is C $5.88\times {10}^{-3}T$

Here, $I=5A;1=0.1m$;
Mass per unit length of the wire
$\lambda =3\times {10}^{-3}kg{m}^{-1}$
Therefore, weight of the wire,
$mg=3\times {10}^{-3}\times 0.1\times 9.8=2.94\times {10}^{-3}N$
Let B be the strength of the magnetic field applied.
$F=BIl=B\times 5\times 0.1=0.5B$
In equilibrium,
$F=mg$
$\therefore 0.5B=2.94\times {10}^{-3}$
$B=\frac{2.94\times {10}^{-3}}{0.5}=5.88\times {10}^{-3}T$