1
JEE Main 2023 (Online) 24th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

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 30 English

A
$$\mathrm{\frac{1}{2}B^2L^2\omega}$$
B
$$\mathrm{\frac{1}{2}BL^2\omega}$$
C
$$\mathrm{\frac{1}{4}BL^2\omega}$$
D
$$\mathrm{\frac{1}{4}B^2L\omega}$$
2
JEE Main 2023 (Online) 24th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

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 :

A
emf = 10 mV
B
emf = 5 mV
C
emf = 100 mV
D
emf = 1 mV
3
JEE Main 2022 (Online) 25th July Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

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

A
$${5 \over 2}$$
B
$${25 \over 4}$$
C
$${5 \over 4}$$
D
$${25 \over 2}$$
4
JEE Main 2022 (Online) 25th July Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A small square loop of wire of side $$l$$ is placed inside a large square loop of wire $$\mathrm{L}(\mathrm{L}>>l)$$. Both loops are coplanar and their centres coincide at point $$\mathrm{O}$$ as shown in figure. The mutual inductance of the system is :

JEE Main 2022 (Online) 25th July Morning Shift Physics - Electromagnetic Induction Question 43 English

A
$$\frac{2 \sqrt{2} \mu_{0} \mathrm{~L}^{2}}{\pi l}$$
B
$$\frac{\mu_{0} l^{2}}{2 \sqrt{2} \pi \mathrm{L}}$$
C
$$\frac{2 \sqrt{2} \mu_{0} l^{2}}{\pi \mathrm{L}}$$
D
$$ \frac{\mu_{0} \mathrm{~L}^{2}}{2 \sqrt{2} \pi l} $$
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