Question

A $250$-Turn rectangular coil of length $2.1cm$ and width $1.25cm$carries a current of $85\mu A$ and subjected to a magnetic field of strength $0.85T$. Work done for rotating the coil by ${180}^o$ against the torque is:

• A. $1.15\mu J$
• B. $9.1\mu J$
• C. $4.55\mu J$
• D. $23\mu J$

Answer 1

Answer: (B)

$9.1\mu J$

$\tau =BINA\sin \theta$
Hence work done
$=\int \tau d\theta$
$BINA{\int }_q^{\pi }\sin \theta d\theta$
$=-BINA(-1-1)$
$=2BINA$
$=2\times 0.85\times 85\times {10}^{-6}\times 250\times (2.1\times 1.25\times {10}^{-5})$
$=94828.125\times {10}^{-10}$
$=9.48\times {10}^{-6}$