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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)  

1 Answer
5876 Votes
3537

Answer Key : (a) -

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 :

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 / 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

(a)

0.75 V

(b)

0.50 V

(c)

0.20 V

(d)

0.25 V


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