The magnetic field of an electromagnetic wave is given by

= 1.6 × 10^{ −6} cos(2 × 10^{7}z + 6 × 10^{15}t)

The associated electric field will be

(a) = 4.8 × 10^{2} cos(2 × 10^{7}z + 6 × 10^{15}t)

(b) = 4.8 × 10^{2} cos(2 × 10^{7}z − 6 × 10^{15}t)

(c) = 4.8 × 10^{2} cos(2 × 10^{7}z − 6 ×10^{15}t)

(d) = 4.8 × 10^{2} cos(2 × 10^{7}z + 6 × 10^{15}t)

The magnetic field of an electromagnetic wave is given by

= 1.6 × 10^{ −6} cos(2 × 10^{7}z + 6 × 10^{15}t)

The associated electric field will be

(a) = 4.8 × 10^{2} cos(2 × 10^{7}z + 6 × 10^{15}t)

(b) = 4.8 × 10^{2} cos(2 × 10^{7}z − 6 × 10^{15}t)

(c) = 4.8 × 10^{2} cos(2 × 10^{7}z − 6 ×10^{15}t)

(d) = 4.8 × 10^{2} cos(2 × 10^{7}z + 6 × 10^{15}t)

1 Answer

5876 Votes

3537

Amplitude of electric field, E = B_{0}C

= 1.6 × 10^{ −6} × × 3 × 10^{8}

= 4.8 × 10^{2} V/m

Also × is along (the direction of propagation)

⇒ = 4.8 × 10^{2} cos(2 × 10^{7}z + 6 × 10^{15}t)

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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