Search any question & find its solution
Question:
Answered & Verified by Expert
A horizontal wire of mass ' $\mathrm{m}$ ', length ' $l$ ' and resistance ' $R$ ' is sliding on the vertical rails on which uniform 'magnetic field ' $B$ ' is directed perpendicular. The terminal speed of the wire as it falls under the force of gravity is ( $\mathrm{g}=$ acceleration due to gravity)
Options:
Solution:
1488 Upvotes
Verified Answer
The correct answer is:
$\frac{\mathrm{mgR}}{\mathrm{B} l}$
Net force on the wire becomes zero when it attains terminal velocity.
$\therefore \quad$ Force due to magnetic field = gravitational force
$\therefore \quad \mathrm{iB} l=\mathrm{mg}$
$\therefore \quad \frac{\mathrm{e}}{\mathrm{R}} \mathrm{B} l=\mathrm{mg} \quad \ldots\left(\because \mathrm{i}=\frac{\mathrm{e}}{\mathrm{R}}\right)$
$\therefore \quad \frac{\mathrm{Bv} l}{\mathrm{R}} \mathrm{B} l=\mathrm{mg} \quad \ldots(\because \mathrm{e}=\mathrm{Bv} l)$
$\therefore \quad \mathrm{v}=\frac{\mathrm{mgR}}{\mathrm{B}^2 l^2}$
$\therefore \quad$ Force due to magnetic field = gravitational force
$\therefore \quad \mathrm{iB} l=\mathrm{mg}$
$\therefore \quad \frac{\mathrm{e}}{\mathrm{R}} \mathrm{B} l=\mathrm{mg} \quad \ldots\left(\because \mathrm{i}=\frac{\mathrm{e}}{\mathrm{R}}\right)$
$\therefore \quad \frac{\mathrm{Bv} l}{\mathrm{R}} \mathrm{B} l=\mathrm{mg} \quad \ldots(\because \mathrm{e}=\mathrm{Bv} l)$
$\therefore \quad \mathrm{v}=\frac{\mathrm{mgR}}{\mathrm{B}^2 l^2}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.