1
JEE Main 2022 (Online) 30th June Morning Shift
MCQ (Single Correct Answer)
+4
-1

An expression for oscillating electric field in a plane electromagnetic wave is given as Ez = 300 sin(5$$\pi$$ $$\times$$ 103x $$-$$ 3$$\pi$$ $$\times$$ 1011t) Vm$$-$$1

Then, the value of magnetic field amplitude will be :

(Given : speed of light in Vacuum c = 3 $$\times$$ 108 ms$$-$$1)

A
1 $$\times$$ 10$$-$$6 T
B
5 $$\times$$ 10$$-$$6 T
C
18 $$\times$$ 109 T
D
21 $$\times$$ 109 T
2
JEE Main 2022 (Online) 30th June Morning Shift
MCQ (Single Correct Answer)
+4
-1

In series RLC resonator, if the self inductance and capacitance become double, the new resonant frequency (f2) and new quality factor (Q2) will be :

(f1 = original resonant frequency, Q1 = original quality factor)

A
$${f_2} = {{{f_1}} \over 2}$$ and $${Q_2} = {Q_1}$$
B
$${f_2} = {f_1}$$ and $${Q_2} = {{{Q_1}} \over {{Q_2}}}$$
C
$${f_2} = 2{f_1}$$ and $${Q_2} = {Q_1}$$
D
$${f_2} = {f_1}$$ and $${Q_2} = 2{Q_1}$$
3
JEE Main 2022 (Online) 29th June Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

For a series LCR circuit, I vs $$\omega$$ curve is shown :

(a) To the left of $$\omega$$r, the circuit is mainly capacitive.

(b) To the left of $$\omega$$r, the circuit is mainly inductive.

(c) At $$\omega$$r, impedance of the circuit is equal to the resistance of the circuit.

(d) At $$\omega$$r, impedance of the circuit is 0.

JEE Main 2022 (Online) 29th June Morning Shift Physics - Alternating Current and Electromagnetic Induction Question 83 English

Choose the most appropriate answer from the options given below :

A
(a) and (d) only.
B
(b) and (d) only.
C
(a) and (c) only.
D
(b) and (c) only.
4
JEE Main 2022 (Online) 28th June Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

An EM wave propagating in x-direction has a wavelength of 8 mm. The electric field vibrating y-direction has maximum magnitude of 60 Vm$$-$$1. Choose the correct equations for electric and magnetic fields if the EM wave is propagating in vacuum :

A

$${E_y} = 60\sin \left[ {{\pi \over 4} \times {{10}^3}(x - 3 \times {{10}^8}t)} \right]\widehat j\,\,V{m^{ - 1}}$$

$${B_z} = 2\sin \left[ {{\pi \over 4} \times {{10}^3}(x - 3 \times {{10}^8}t)} \right]\widehat k\,\,T$$

B

$${E_y} = 60\sin \left[ {{\pi \over 4} \times {{10}^3}(x - 3 \times {{10}^8}t)} \right]\widehat j\,\,V{m^{ - 1}}$$

$${B_z} = 2 \times {10^{ - 7}}\sin \left[ {{\pi \over 4} \times {{10}^3}(x - 3 \times {{10}^8}t)} \right]\widehat k\,\,T$$

C

$${E_y} = 2 \times {10^{ - 7}}\sin \left[ {{\pi \over 4} \times {{10}^3}(x - 3 \times {{10}^8}t)} \right]\widehat j\,\,V{m^{ - 1}}$$

$${B_z} = 60\sin \left[ {{\pi \over 4} \times {{10}^3}(x - 3 \times {{10}^8}t)} \right]\widehat k\,\,T$$

D

$${E_y} = 2 \times {10^{ - 7}}\sin \left[ {{\pi \over 4} \times {{10}^4}(x - 4 \times {{10}^8}t)} \right]\widehat j\,\,V{m^{ - 1}}$$

$${B_z} = 60\sin \left[ {{\pi \over 4} \times {{10}^4}(x - 4 \times {{10}^8}t)} \right]\widehat k\,\,T$$

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