JEE Main 2023 (Online) 25th January Evening Shift | Electromagnetic Induction Question 44 | Physics

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 39 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

JEE Main 2022 (Online) 25th July Evening Shift

MCQ (Single Correct Answer)

-1

The electric current in a circular coil of 2 turns produces a magnetic induction B₁ at its centre. The coil is unwound and in rewound into a circular coil of 5 tuns and the same current produces a magnetic induction B₂ at its centre. The ratio of $${{{B_2}} \over {{B_1}}}$$ is

$${5 \over 2}$$

$${25 \over 4}$$

$${5 \over 4}$$

$${25 \over 2}$$

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