Question

A rectangular loop of sides 6 cm and 2 cm with a small cut is moving Out of a region of uniform magnetic field of magnitude 0.4 T directed normal to the loop. The voltage developed across the cut if velocity of loop is 2 cm s in a direction normal to the longer side is then

• A. $1.6\times {10}^{-4}$
• B. $2.2\times {10}^{-2}$
• C. $3.2\times {10}^{-2}$
• D. $3.8\times {10}^{-4}$

Answer 1

Answer: (A)

Here, $l=6cm=6\times {10}^{-2}m{l}^{\prime }=2$

$B=0.4$ T,

$v=2cm{s}^{-1}$

$=2\times {10}^{-2}m{s}^{-1}$

Voltage developed across tthe cut is equal to the voltage develop in the rod of length equal to cut$l^{\prime }$.

$\epsilon =Blv=0.4\times 2\times {10}^{-2}\times 2\times {10}^{-2}$

$\epsilon =1.6\times {10}^{-4}V$