Last Updated On [03-11-2023]
The magnetic field of an electromagnetic wave is given by
= 1.6 × 10 −6 cos(2 × 107z + 6 × 1015t)
The associated electric field will be
(a) 
= 4.8 × 102 cos(2 × 107z + 6 × 1015t)
(b) 
= 4.8 × 102 cos(2 × 107z − 6 × 1015t) 
(c) 
= 4.8 × 102 cos(2 × 107z − 6 ×1015t)
(d) 
= 4.8 × 102 cos(2 × 107z + 6 × 1015t) 
Question ID - 1 :- The magnetic field of an electromagnetic wave is given by
= 1.6 × 10 −6 cos(2 × 107z + 6 × 1015t)
The associated electric field will be
(a) = 4.8 × 102 cos(2 × 107z + 6 × 1015t)
(b) = 4.8 × 102 cos(2 × 107z − 6 × 1015t)
(c) = 4.8 × 102 cos(2 × 107z − 6 ×1015t)
(d) = 4.8 × 102 cos(2 × 107z + 6 × 1015t)
The magnetic field of an electromagnetic wave is given by
= 1.6 × 10 −6 cos(2 × 107z + 6 × 1015t)
The associated electric field will be
(a) = 4.8 × 102 cos(2 × 107z + 6 × 1015t)
(b) = 4.8 × 102 cos(2 × 107z − 6 × 1015t)
(c) = 4.8 × 102 cos(2 × 107z − 6 ×1015t)
(d) = 4.8 × 102 cos(2 × 107z + 6 × 1015t)
Amplitude of electric field, E = B0C
= 1.6 × 10 −6 × 
× 3 × 108
= 4.8 × 102
V/m
Also 
× 
is along 
(the direction of propagation)
⇒ 
= 4.8 × 102 cos(2 × 107z + 6 × 1015t)
Next Question :
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In the circuit shown, a four −wire potentiometer is made of a 400 cm long wire, which extends between A and B. The resistance per unit length of the potentiometer wire is r = 0.01
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(a) |
0.75 V |
(b) |
0.50 V |
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(c) |
0.20 V |
(d) |
0.25 V |
/ cm. If an ideal voltmeter is connected as shown with jockey J at 50 cm from end A, the expected reading of the voltmeter will be
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