JEE Main 2019 (Online) 10th January Morning Slot | Alternating Current and Electromagnetic Induction Question 170 | Physics

JEE Main 2019 (Online) 10th January Morning Slot

MCQ (Single Correct Answer)

-1

If the magnetic field of a plane electromagnetic wave is given by (the speed of light = 3 × 10⁸ B = 100 × 10^–6 sin $$\left[ {2\pi \times 2 \times {{10}^{15}}\left( {t - {x \over c}} \right)} \right]$$ then the maximum electric field associated with it is -

4.5 $$ \times $$ 10⁴ N/C

4 $$ \times $$ 10⁴ N/C

6 $$ \times $$ 10⁴ N/C

3 $$ \times $$ 10⁴ N/C

JEE Main 2019 (Online) 9th January Evening Slot

MCQ (Single Correct Answer)

-1

A power transmission line feeds input power at 2300 V to a srep down transformer with its primary windings having 4000 turns. The output power is delivered at 230 V by the transformer. If the current in the primary of the transformer is 5A and its efficiency is 90%, the output current would be :

50 A

45 A

35 A

25 A

JEE Main 2019 (Online) 9th January Evening Slot

MCQ (Single Correct Answer)

-1

A series AC circuit containing an inductor (20 mH), a capacitor (120 $$\mu $$F) and a resistor (60 $$\Omega $$) is driven by an AC source of 24V/50 Hz. The energy dissipated in the circuit in 60 s is :

5.65 $$ \times $$ 10²J

2.26 $$ \times $$ 10³J

5.17 $$ \times $$ 10² J

3.39 $$ \times $$ 10³ J

JEE Main 2018 (Online) 16th April Morning Slot

MCQ (Single Correct Answer)

-1

A coil of cross-sectional area A having n turns is placed in a uniform magnetic field B. When it is rotated with an angular velocity $$\omega ,$$ the maxium e.m.f. induced in the coil will be:

3 nBA$$\omega $$

$${3 \over 2}$$ nBA$$\omega $$

nBA$$\omega $$

$${1 \over 2}$$ nBA$$\omega $$

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