KVPY SX SB Exam 19-Nov-2017 Question Paper

Question : 94 of 120

Marks: +1, -0

A square-shaped wire loop of mass m, resistance R and side 'a' moving with speed

v_0

, parallel to the x-axis, enters a region of uniform magnetic field B, which is perpendicular to the plane of the loop. The speed of the loop changes with distance x (x < a) in the filed, as

$v_0 -{B ^2a ^2}/{Rm}x$
$v_0 -{B ^2a ^2}/{2Rm}x$
$v_0 -{B ^2a ^2}/{Rm}x^2$
$v_0$

Validate

Solution:

i = {vBa}/R

F = ilB ={vBa .Ba}R

ma={vB ^2a ^2}/{R }

{-mdv}/{dt}={vB ^2a ^2}/{R }

⇒

-m{vdv}/{dt}=v{B ^2a ^2}/{R }

-m∫↖{v}↙{v_0}dv={B ^2a ^2}/{R } ∫↖{x}↙{0}dx

-m (v -v_0) = {B ^2a ^2x}/{R }

⇒

v = v_0- {B ^2a ^2}/{mR }x

Go to Question:

Prev Question

Next Question