MHT CET 2023 10th May Morning Shift | Electromagnetic Induction Question 63 | Physics

A square loop of area $$25 \mathrm{~cm}^2$$ has a resistance of $$10 \Omega$$. This loop is placed in a uniform magnetic field of magnitude $$40 \mathrm{~T}$$. 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 one second, will be

$$1 \times 10^{-4} \mathrm{~J}$$

$$1.0 \times 10^{-3} \mathrm{~J}$$

$$5 \times 10^{-3} \mathrm{~J}$$

$$2.5 \times 10^{-3} \mathrm{~J}$$

MHT CET 2023 10th May Morning Shift

MCQ (Single Correct Answer)

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Two conducting circular loops of radii '$$R_1$$' and '$$R_2$$' are placed in the same plane with their centres coinciding. If $$R_1>R_2$$, the mutual inductance $$M$$ between them will be directly proportional to

$$\frac{R_1}{R_2}$$

$$\frac{R_2}{R_1}$$

$$\frac{R_1^2}{R_2}$$

$$\frac{R_2^2}{R_1}$$

MHT CET 2023 9th May Evening Shift

MCQ (Single Correct Answer)

-0

If current '$$I$$' is flowing in the closed circuit with collective resistance '$$R$$', the rate of production of heat energy in the loop as we pull it along with a constant speed '$$\mathrm{V}$$' is ( $$\mathrm{L}=$$ length of conductor, $$\mathrm{B}=$$ magnetic field)

$$\frac{\mathrm{BLV}}{\mathrm{R}}$$

$$\frac{\mathrm{B}^2 \mathrm{~L}^2 \mathrm{~V}^2}{\mathrm{R}^2}$$

$$\frac{B L V}{R^2}$$

$$\frac{\mathrm{B}^2 \mathrm{~L}^2 \mathrm{~V}^2}{\mathrm{R}}$$

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