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xz_007 [3.2K]
3 years ago
6

A current carrying loop of width a and length b is placed near a current carrying wire. How does the net force on the loop compa

re to the net force on a single wire segment of length a carrying the same amount of current placed at the same distance from the wire
Physics
1 answer:
makvit [3.9K]3 years ago
5 0

Answer/ Explanation

the loop has no force and the wires cancel each other but the wire on the loop has a force

The segment a get the same force for both but in the loop segment b get an opposite force so the net force on the loop is smaller

The loop wire has forces that all cancel out while the other straight wire doesn't.

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Answer:

V_d = 1.75 × 10⁻⁴ m/s

Explanation:

Given:

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Current, I = 3A

Area of the wire, A = \frac{\pi d^2}{4} = A = \frac{\pi 0.625^2}{4}

Now,

The current density, J is given as

J=\frac{I}{A}=\frac{3}{ \frac{\pi 0.625^2}{4}}= 2444619.925 A/mm²

now, the electron density, n = \frac{\rho}{M}N_A

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N_A=Avogadro's Number

n = \frac{8.93}{63.5}(6.2\times 10^{23})=8.719\times 10^{28}\ electrons/m^3

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V_d=\frac{J}{ne}

where,

e = charge on electron = 1.6 × 10⁻¹⁹ C

thus,

V_d=\frac{2444619.925}{8.719\times 10^{28}\times (1.6\times 10^{-19})e} = 1.75 × 10⁻⁴ m/s

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