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Aloiza [94]
3 years ago
15

Two parallel rods are each 0.69 m in length. They are attached at their centers to a spring that is initially neither stretched

nor compressed. The spring has a spring constant of 130 N/m. When 1200 A of current is in each rod in the same direction, the spring is observed to be compressed by 3.0 cm. Treat the rods as long, straight wires and find the separation between them when the current is present.
Physics
1 answer:
diamong [38]3 years ago
5 0

Answer:

Explanation:

Let the separation required be d .

Force between rod = 10⁻⁷ x  2 I₁ I₂ L / d

where I₁ and I₂ are current in them , d is distance of separation and L is length of wire .

Force between rod = 10⁻⁷ x  2 x 1200 x  1200 x .69  / d

= .1987 /d

Restoring Force by spring = k x where k is force constant and x is compression .

= 130 x .03

= 3.9 N

For balancing

Restoring Force by spring = Force between rod

.1987 /d = 3.9

d = .1987 /3.9

= .0509 m

= 5.09 cm .

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To find :-

find the induced Emf if rate of change of the magnetic field and the number of turns in the coil are Doubled (but nothing else changes).

So,

   Emf induced in the coil represented by formula

                          ∈  =   -N\frac{d\phi}{dt}                                  ...................(1)

                                          Where:

                                                    .   \phi = BAcos\theta     { B is magnetic field }

                                                                                 {A is cross-sectional area}

                                                    .  N = No. of turns in coil.

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Here,

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