A 7.0 m ordinary lamp extension cord carries a current, I = 7.0 A. Such a cord typically consists of two parallel wires carrying equal currents in opposite directions. The two wires are separated by about 4 mm. Find the magnitude and direction of the force that the two segments of this cord exert on each other. A) 2 N , The wires repel B) 2 N, The wires attract C) 0.02 N , The wires repel D) 0.02 N, The wires attract

(C) , The wires repel

Explanation:

We are given that

Length of wire=L=7 m

We have to find the magnitude and direction of the force that the two segments of this cord exert on each other.

Where

Using the formula

When the wires carrying current in equal current in  opposite direction then, the wire repel to each other.

Hence, the wires repel to each other.

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energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j