1
MCQ (Single Correct Answer)

JEE Main 2017 (Online) 9th April Morning Slot

The electric field component of a monochromatic radiation is given by

$$\overrightarrow E $$ = 2 E0 $$\widehat i$$ cos kz cos $$\omega $$t

Its magnetic field $$\overrightarrow B $$ is then given by :
A
$${{2{E_0}} \over c}$$ $$\widehat j$$ sin kz cos $$\omega $$t
B
$$-$$ $${{2{E_0}} \over c}$$ $$\widehat j$$ sin kz sin $$\omega $$t
C
$${{2{E_0}} \over c}$$ $$\widehat j$$ sin kz sin $$\omega $$t
D
$${{2{E_0}} \over c}$$ $$\widehat j$$ cos kz cos $$\omega $$t
2
MCQ (Single Correct Answer)

JEE Main 2018 (Offline)

In an a.c. circuit, the instantaneous e.m.f. and current are given by
e = 100 sin 30 t
i = 20 sin $$\left( {30t - {\pi \over 4}} \right)$$
In one cycle of a.c., the average power consumed by the circuit and the wattless current are, respectively
A
50, 0
B
50, 10
C
$${{1000} \over {\sqrt 2 }},10$$
D
$${{50} \over {\sqrt 2 }}$$

Explanation

Wattless current,

here  $$\phi $$  is the angle between i and e.

Average power,

Pav = Vrms Irms cos$$\phi $$

= $${{100} \over {\sqrt 2 }} \times {{20} \over {\sqrt 2 }}$$ cos$${\pi \over 4}$$

= $${{1000} \over {\sqrt 2 }}$$ watt.
3
MCQ (Single Correct Answer)

JEE Main 2018 (Offline)

For an RLC circuit driven with voltage of amplitude vm and frequency $${\omega _0}$$ = $${1 \over {\sqrt {LC} }}$$ the current exhibits resonance. The quality factor, Q is given by :
A
$${{CR} \over {{\omega _0}}}$$
B
$${{{\omega _0}L} \over R}$$
C
$${{{\omega _0}R} \over L}$$
D
$${R \over {\left( {{\omega _0}C} \right)}}$$

Explanation

Quality factor (Q) = $${{Angular\,\,{\mathop{\rm Re}\nolimits} sonance} \over {Bandwith}}$$

= $${{{1 \over {\sqrt {LC} }}} \over {{R \over L}}}$$

= $${{{\omega _0}} \over {{R \over L}}}$$

= $${{{\omega _0}L} \over R}$$
4
MCQ (Single Correct Answer)

JEE Main 2018 (Offline)

An EM wave from air enters a medium. The electric fields are
$$\overrightarrow {{E_1}} $$ = $${E_{01}}\widehat x\cos \left[ {2\pi v\left( {{z \over c} - t} \right)} \right]$$ in air and $$\overrightarrow {{E_2}} $$ = $${E_{02}}\widehat x\cos \left[ {k\left( {2z - ct} \right)} \right]$$ in medium, where the wave number k and frequency $$\nu $$ refer to their values in air. The medium is non-magnetic. If $${\varepsilon _{{r_1}}}$$ and $${\varepsilon _{{r_2}}}$$ refer to relative permittivities of air and medium respectively, which of the following options is correct ?
A
$${{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = 4$$
B
$${{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = 2$$
C
$${{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = {1 \over 4}$$
D
$${{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = {1 \over 2}$$

Explanation

Electric field in air,

$$\overrightarrow {{E_1}} $$  =  E01 $$\widehat x$$ cos ( $${{2\pi vz} \over c}$$ $$-$$ 2$$\pi $$vt )

$$\therefore\,\,\,$$ Velocity in air = $${{2\pi v} \over {{{2\pi v} \over c}}}$$ = c

Also,    c = $${1 \over {\sqrt {\mu \varepsilon {r_1}{\varepsilon _0}} }}$$ . . . . . . (1)

$$\overrightarrow {{E_2}} $$  =  E02 $$\widehat x$$ cos(2kz $$-$$ kct)

$$\therefore\,\,\,$$ Velocity in medium = $${{kc} \over {2k}}$$ = $${c \over 2}$$

Also, $${c \over 2}$$ = $${1 \over {\sqrt {\mu {\varepsilon _{r2}}\,{\varepsilon _0}} }}$$ . . . . . (2)

As, medium is non magnetic,

So,   $$\mu $$medium = $$\mu $$air = $$\mu $$

Dividing (1) by (2), we get

2 = $$\sqrt {{{{\varepsilon _{r2}}} \over {{\varepsilon _{r1}}}}} $$

$$ \Rightarrow $$$$\,\,\,$$ $${{{{\varepsilon _{r1}}} \over {{\varepsilon _{r2}}}}}$$ = $${1 \over 4}$$

Questions Asked from Alternating Current and Electromagnetic Induction

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