### 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

## Show Answer

#### Answer: (2)

#### Solution:

#### Formula: Magnetic Field Due To Infinite Straight Wire

$\ell=50 cm$

$t=1 sec$

$\therefore V=\frac{0.05}{1}=0.05 m / s$.

$i=\frac{40 \times 0.05 \times 0.05}{10}=0.01 A$

$F=B_i \ell=40 \times 0.01 \times 0.05$

$F=0.02 N$

$\therefore \quad W=0.02 \times \ell=0.02 \times .05$

$\therefore \quad W=1 \times 10^{-3} J$