Question

A straight wire of mass $200\, g$ and length $1.5\, m$ carries a current of $2\, A$. It is suspended in mid-air by a uniform horizontal magnetic field $B$. The magnitude of $B$ (in tesla) is (assume $g=9.8\, ms ^{-2}$ )

• A. 2
• B. 1.5
• C. 0.55
• D. 0.65

Answer 1

Answer: (D)

Magnetic force on straight wire
$F=$ Bil $\sin \theta=$ Bil $\sin 90^{\circ}=$ Bil
For equilibrium of wire in mid-air,
$F =m g$
$Bil =m g$
$\therefore B=\frac{m g}{il}=\frac{200 \times 10^{-3} \times 9.8}{2 \times 1.5}$
$=0.65\, T$