Magnetic Effects Of Current Question 11
Question 11 - 29 January - Shift 2
A square loop of area $25 cm^{2}$ has a resistance of $10 \Omega$. The loop is placed in uniform magnetic field of magnitude $40.0 T$. The plane of loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in $1.0 sec$, will be
(1) $2.5 \times 10^{-3} J$
(2) $1.0 \times 10^{-3} J$
(3) $1.0 \times 10^{-4} J$
(4) $5 \times 10^{-3} J$
nan
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Answer: (2)
Solution:
Formula: Magnetic Field Due To Infinite Straight Wire
$\ell=50 cm$
$t=1 sec$
$\therefore V=\frac{0.05}{1}=0.05 m / s$.
$i=\frac{40 \times 0.05 \times 0.05}{10}=0.01 A$
$F=B_i \ell=40 \times 0.01 \times 0.05$
$F=0.02 N$
$\therefore \quad W=0.02 \times \ell=0.02 \times .05$
$\therefore \quad W=1 \times 10^{-3} J$
