A square loop of area 25 cm ^ 2 has a resistance of 10Ω . The loop...

As ${l}^{2} = 25{cm}^{2}$, then $l = 5cm = 0 . 05m$.

Given: $t = 1s$

Velocity of the square loop, $v = \dfrac{0 . 05}{1} = 0 . 05m{s}^{- 1}$.

Induced current,

$i = \dfrac{V}{R} = \dfrac{Blv}{R} = \dfrac{40 \times 0 . 05 \times 0 . 05}{10} = 0 . 01A$

Now force acting on the side of the square loop, $F = Bil = 40 \times 0 . 01 \times 0 . 05$

$⇒ F = 0 . 02N$

Therefore, work done

$W = Fl = 0 . 02 \times l = 0 . 02 \times 0 . 05$

$⇒ W = 1 \times {10}^{- 3}J$