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

Solution :

Given :

Area of square loop,

$$A = 25\text{ cm}^2$$

$$= 25 \times 10^{-4}\text{ m}^2$$

Resistance of loop,

$$R = 10\ \Omega$$

Magnetic field,

$$B = 40\text{ T}$$

Time taken,

$$t = 1\text{ s}$$

Initial magnetic flux through the loop :

$$\phi_i = BA$$

Final magnetic flux after removing the loop :

$$\phi_f = 0$$

Therefore,

$$\Delta \phi = BA$$

Induced emf :

$$\mathcal{E} = \frac{\Delta \phi}{t}$$

$$= \frac{BA}{t}$$

Substituting values :

$$\mathcal{E}=\frac{40 \times 25 \times 10^{-4}}{1}$$

$$=\frac{1000}{10^4}$$

$$=0.1\text{ V}$$

Induced current :

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

$$= \frac{0.1}{10}$$

$$= 0.01\text{ A}$$

Work done in pulling the loop slowly equals heat produced :

$$W = I^2Rt$$

$$= (0.01)^2 \times 10 \times 1$$

$$= 10^{-3}\text{ J}$$

Final Answer :

$$10^{-3}\text{ J}$$