To calculate the work done in pulling the loop out of the magnetic field, we can use the induced electromotive force (emf) and subsequently the power and energy dissipated due to the resistance.

The change in magnetic flux, $\Delta \Phi_B$, as the loop is pulled completely out of the magnetic field is given by the initial magnetic flux, since the final magnetic flux will be zero. The magnetic flux through the loop is:

$$ \Phi_B = B \times A $$

where:

$B = 40 \, \text{T}$ is the magnetic field strength,

$A = 25 \, \text{cm}^2 = 25 \times 10^{-4} \, \text{m}^2 = 0.0025 \, \text{m}^2$ is the area of the square loop.

Thus, the initial magnetic flux is:

$$ \Phi_B = 40 \times 0.0025 = 0.1 \, \text{Wb} $$

The change in magnetic flux is:

$$ \Delta \Phi_B = 0.1 \, \text{Wb} $$

According to Faraday's law of electromagnetic induction, the induced emf, $\mathcal{E}$, is:

$$ \mathcal{E} = -\frac{\Delta \Phi_B}{\Delta t} $$

Since we are pulling the loop out in a time interval of $\Delta t = 1 \, \text{s}$:

$$ \mathcal{E} = -\frac{0.1}{1} = -0.1 \, \text{V} $$

The negative sign indicates the direction of the induced emf, but for the magnitude of work done, we can consider the positive value due to the absolute work calculation.

The current induced in the loop, $I$, can be calculated using Ohm's Law as:

$$ I = \frac{\mathcal{E}}{R} $$

where $R = 10 \, \Omega$ is the resistance of the loop:

$$ I = \frac{0.1}{10} = 0.01 \, \text{A} $$

The power dissipated as heat in the resistor due to this current is:

$$ P = I^2 R = (0.01)^2 \times 10 = 0.001 \, \text{W} $$

Since the loop is being pulled out over a time period of $1 \, \text{s}$, the total work done (energy dissipated) is:

$$ W = P \times \Delta t = 0.001 \times 1 = 0.001 \, \text{J} = 1.0 \times 10^{-3} \, \text{J} $$

Therefore, the work done in pulling the loop out of the magnetic field is option B: $1.0 \times 10^{-3} \, \text{J}$.

Electromagnetic Induction

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