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kobusy [5.1K]
4 years ago
9

A copper wire has radius 0.800 mm and carries current I at 20.0°C. A silver wire with radius 0.500 mm carries the same current a

nd is also at 20.0°C. Let Ecu be the electric field in the copper wire and Eag be the electric field in the silver wire. Use the resistivities at room temperature for copper. PCu = 1.72 x 10^-8.m and for silver PAE= 1.47 x 10^-8.m. What is the ratio Ecu/Eng?
Physics
1 answer:
IgorC [24]4 years ago
7 0

Answer:

Ecu/Eag = 0.46

Explanation:

E = PI/A

Ecu = Pcu × I/A

Pcu = 1.72×10^-8 ohm-meter

r = 0.8 mm = 0.8/1000 = 8×10^-4 m

A = πr^2 = π×(8×10^-4)^2 = 6.4×10^-7π

Ecu = 1.72×10^-8I/6.4×10^-7π = 0.026875I/1

Eag = Pag × I/A

Pag = 1.47×10^-8 ohm-meter

r = 0.5 mm = 0.5/1000 = 5×10^-4 m

A = πr^2 = π × (5×10^-4)^2 = 2.5×10^-7π

Eag = 1.47×10^-8I/2.5×10^-7π = 0.0588I/π

Ecu/Eag = 0.026875I/π × π/0.0588I = 0.46

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Karo-lina-s [1.5K]

Answer:

44 N

Explanation:

The electrostatic forces between two charges is given by:

F=k\frac{q_1 q_2}{r^2}

where

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r is their separation

We notice that the force is directly proportional to the charges.

In this problem, initially we have a force of

F = 22 N

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If the charge is doubled,

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A. 6.13 x 107

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3 years ago
Albert's laboratory is filled with a constant uniform magnetic field pointing straight up. Albert throws some charges into this
guajiro [1.7K]

Answer:

\vec{F}=qB(v_y \hat{i} - v_x\hat{j})

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Lets consider the z-direction of our coordinate system the same direction of the magnetic field, that is:

\vec{B} = B \hat{k}

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