Question

A wire 40cm long is bent into a rectangular frame 15 cm $\times$ 5 cm and placed perpendicular to a field of induction 0.8 T in 0.5 sec, the frame is changed into a square frame and the field is increased to 1.4T The e.m.f induced in the frame

• A. $1.6\times {10}^{-2}V$
• B. $1.6\times {10}^{-3}V$
• C. $1.6\times {10}^{-4}V$
• D. $1.6\times {10}^{-5}V$

Answer 1

The correct option is A $1.6\times {10}^{-2}V$

$A_1=15\times 5=75c{m}^2$
$\begin{array}{ccc}{A}_2=a^2& |& 4a=40\\ =10\times 10& & a=10cm\\ A_2=100c{m}^2& & \end{array}$
$e=\frac{B_2{A}_2-B_1{A}_1}{t}=\frac{(1.4\times 100-0.8\times 75){10}^{-4}}{0.5}$
$=1.6\times {10}^{-2}V$