JEE Main 2019 (Online) 11th January Morning Slot | Alternating Current and Electromagnetic Induction Question 167 | Physics

JEE Main 2019 (Online) 11th January Morning Slot

MCQ (Single Correct Answer)

-1

An electromagnetic wave of intensity 50 Wm^–2 enters in a medium of refractive index 'n' without any loss. The ratio of the magnitudes of electric, and the ratio of the magnitudes of magnetic fields of the wave before and after entering into the medium are respectively, given by:

$$\left( {{1 \over {\sqrt n }},{1 \over {\sqrt n }}} \right)$$

$$\left( {\sqrt n ,\sqrt n } \right)$$

$$\left( {\sqrt n ,{1 \over {\sqrt n }}} \right)$$

$$\left( {{1 \over {\sqrt n }},\sqrt n } \right)$$

JEE Main 2019 (Online) 10th January Evening Slot

MCQ (Single Correct Answer)

-1

The self induced emf of a coil is 25 volts. When the current in it is changed at uniform rate from 10 A to 25 A in 1s, the change in the energy of the inductance is -

740 J

637.5 J

540 J

437.5 J

JEE Main 2019 (Online) 10th January Evening Slot

MCQ (Single Correct Answer)

-1

The electric field of a plane polarized electromagnetic wave in free space at time t = 0 is given by an expression $$\overrightarrow E \left( {x,y} \right) = 10\widehat j\cos \left[ {\left( {6x + 8z} \right)} \right].$$ The magnetic field $$\overrightarrow B $$(x,z, t) is given by $$-$$ (c is the velocity of light)

$${1 \over c}\left( {6\hat k + 8\widehat i} \right)\cos \left[ {\left( {6x + 8z - 10ct} \right)} \right]$$

$${1 \over c}\left( {6\widehat k - 8\widehat i} \right)\cos \left[ {\left( {6x + 8z - 10ct} \right)} \right]$$

$${1 \over c}\left( {6\hat k + 8\widehat i} \right)\cos \left[ {\left( {6x - 8z + 10ct} \right)} \right]$$

$${1 \over c}\left( {6\hat k - 8\widehat i} \right)\cos \left[ {\left( {6x - 8z + 10ct} \right)} \right]$$

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

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