Q. 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
A) 2
B) 1.5
C) 0.55
D) 0.65
Explanation
(d) Magnetic force on straight wire
$F = Bil\sin\theta$
$\ \ \ \ \ \ \ \ = Bil\sin{90{^\circ}} = Bil$
For equilibrium of wire in mid-air,
$\ \ \ \ F = mg$
$\ Bil = mg$
$\therefore B = \frac{mg}{il}$
$\ \ \ \ \ \ \ \ = \frac{200 \times 10^{- 3} \times 9.8}{2 \times 1.5} = 0.65\ T$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.