JEE Main 2023 (Online) 29th January Morning Shift | Electromagnetic Induction Question 44 | Physics

Find the mutual inductance in the arrangement, when a small circular loop of wire of radius '$$R$$' is placed inside a large square loop of wire of side $$L$$ $$(L \gg R)$$. The loops are coplanar and their centres coincide :

JEE Main 2023 (Online) 29th January Morning Shift Physics - Electromagnetic Induction Question 44 English

$$M=\frac{\sqrt{2} \mu_{0} R}{L^{2}}$$

$$M=\frac{2 \sqrt{2} \mu_{0} R}{L^{2}}$$

$$M=\frac{2 \sqrt{2} \mu_{0} R^{2}}{L}$$

$$M=\frac{\sqrt{2} \mu_{0} R^{2}}{L}$$

JEE Main 2023 (Online) 25th January Evening Shift

MCQ (Single Correct Answer)

-1

A wire of length 1m moving with velocity 8 m/s at right angles to a magnetic field of 2T. The magnitude of induced emf, between the ends of wire will be __________.

20 V

8 V

16 V

12 V

JEE Main 2023 (Online) 24th January Evening Shift

MCQ (Single Correct Answer)

-1

A metallic rod of length 'L' is rotated with an angular speed of '$$\omega$$' normal to a uniform magnetic field 'B' about an axis passing through one end of rod as shown in figure. The induced emf will be :

JEE Main 2023 (Online) 24th January Evening Shift Physics - Electromagnetic Induction Question 41 English

$$\mathrm{\frac{1}{2}B^2L^2\omega}$$

$$\mathrm{\frac{1}{2}BL^2\omega}$$

$$\mathrm{\frac{1}{4}BL^2\omega}$$

$$\mathrm{\frac{1}{4}B^2L\omega}$$

JEE Main 2023 (Online) 24th January Morning Shift

MCQ (Single Correct Answer)

-1

A conducting circular loop of radius $$\frac{10}{\sqrt\pi}$$ cm is placed perpendicular to a uniform magnetic field of 0.5 T. The magnetic field is decreased to zero in 0.5 s at a steady rate. The induced emf in the circular loop at 0.25 s is :

emf = 10 mV

emf = 5 mV

emf = 100 mV

emf = 1 mV

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