JEE Main 2020 (Online) 3rd September Morning Slot | Alternating Current and Electromagnetic Induction Question 139 | Physics

JEE Main 2020 (Online) 3rd September Morning Slot

MCQ (Single Correct Answer)

-1

The magnetic field of a plane electromagnetic wave is
$$\overrightarrow B = 3 \times {10^{ - 8}}\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat i$$ T
where c = 3 $$ \times $$ 10⁸ ms^–1 is the speed of light. The corresponding electric field is :

$$\overrightarrow E = - {10^{ - 6}}\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k$$ V/m

$$\overrightarrow E = - 9\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k$$ V/m

$$\overrightarrow E = 9\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k$$ V/m

$$\overrightarrow E = 3 \times {10^{ - 8}}\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k$$

JEE Main 2020 (Online) 3rd September Morning Slot

MCQ (Single Correct Answer)

-1

A 750 Hz, 20 V (rms) source is connected to a resistance of 100 $$\Omega $$, an inductance of 0.1803 H and a capacitance of 10 $$\mu $$F all in series. The time in which the resistance (heat capacity 2 J/^oC) will get heated by 10^oC. (assume no loss of heat to the surroudnings) is close to :

348 s

418 s

245 s

365 s

JEE Main 2020 (Online) 2nd September Evening Slot

MCQ (Single Correct Answer)

-1

In a plane electromagnetic wave, the directions of electric field and magnetic field are represented by $$\widehat k$$ and $$2\widehat i - 2\widehat j$$, respectively. What is the unit vector along direction of propagation of the wave?

$${1 \over {\sqrt 5 }}\left( {\widehat i + 2\widehat j} \right)$$

$${1 \over {\sqrt 5 }}\left( {2\widehat i + \widehat j} \right)$$

$${1 \over {\sqrt 2 }}\left( {\widehat i + \widehat j} \right)$$

$${1 \over {\sqrt 2 }}\left( {\widehat j + \widehat k} \right)$$

JEE Main 2020 (Online) 2nd September Evening Slot

MCQ (Single Correct Answer)

-1

An inductance coil has a reactance of 100 $$\Omega $$. When an AC signal of frequency 1000 Hz is applied to the coil, the applied voltage leads the current by 45^o. The self-inductance of the coil is

6.7 $$ \times $$ 10^–7 H

1.1 $$ \times $$ 10^–1 H

5.5 $$ \times $$ 10^–5 H

1.1 $$ \times $$ 10^–2 H

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