Q. A wire of length 50 cm moves with a velocity of 300 m-mi$n^{- 1},$ perpendicular to a magnetic field. If the emf induced in the wire is 2 V, the magnitude of the field in telsa is
A) 2
B) 5
C) 0.4
D) 0.8
Explanation
(d) Emf induced in the wire is given by
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ e = Blv$
Given, $\ \ \ \ \ \ \ \ \ l = 50\ cm = 0.5\ m$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v = 300\ m -$min$^{-1}$=$\frac{300}{60} = 5\ {ms}^{- 1}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ e = 2\ V$
Magnetic field, $B = \frac{e}{lv} = \frac{2}{0.5 \times 5} = 0.8\ T$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.