Question

A straight wire of mass 200 g and length 1.5 m carries a current of 2 A . It is suspending in mid-air by a uniform horizontal magnetic filed B. The magnitude of B (in tesla) is

• A. $2$
• B. $1.5$
• C. $0.55$
• D. $0.65$

Answer 1

The correct option is D $0.65$

Given: Mass of the wire = 200 g

Length of the wire = 1.5 m

Current in the wire = 2A

Solution: As we know, force acting on a straight wire due to the magnetic field is given by,

$F=BIl\sin \theta =BIl\sin {90}^{\circ }$

$=BIl$

According to the question,

for the wire to be in equilibrium in mid air,

$F=mg$

$BIl=mg$

$\therefore B=\frac{mg}{Il}$

$=\frac{200\times {10}^{-3}\times 9.8}{2\times 1.5}$

$=0.65T$

Hence, the correct option is (D).