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Harman [31]
3 years ago
11

Two long parallel wires carry currents of 3.05 A 3.05 A and 6.69 A 6.69 A . The magnitude of the force per unit length acting on

each wire is 7.33 × 10 − 5 N / m 7.33×10−5 N/m . Find the separation distance d d of the wires expressed in millimeters.
Physics
1 answer:
ziro4ka [17]3 years ago
7 0

Answer:

The separation between wires is 55.6 mm.

Explanation:

Given that,

Current in wire 1, I_1=3.05\ A

Current in wire 2, I_2=6.69\ A

The magnitude of the force per unit length acting on each wire is, \dfrac{F}{l}=7.33\times 10^{-5}\ N/m

We know that the formula of magnitude of the force per unit length is given by :

\dfrac{F}{l}=\dfrac{\mu_oI_1I_2}{2\pi d}

Here, d is the separation between wires.

d=\dfrac{\mu_o I_1I_2}{2\pi (F/l)}\\\\d=\dfrac{4\pi \times 10^{-7}\times 3.05\times 6.69}{2\pi \times 7.33\times 10^{-5}}\\\\d=0.0556\ m\\\\d=55.6\ mm

So, the separation between wires is 55.6 mm.

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3 years ago
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Force → F = ( − 8.0 N ) ˆ i + ( 6.0 N ) ˆ j acts on a particle with position vector → r = ( 3.0 m ) ˆ i + ( 4.0 m ) ˆ j . What a
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Explanation:

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Force, F=(-8\ N)i+(6\ N)j

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(b) To find the angle between r and F use dot product formula as :

F{\cdot} r=|F||r|\ \cos\theta\\\\\cos\theta=\dfrac{F{\cdot} r}{|F| |r|}\\\\\cos\theta=\dfrac{(-8i+6j){\cdot} (3i+4j)}{\sqrt{(-8)^2+6^2} \times \sqrt{3^2+4^2} }\\\\\cos\theta=\dfrac{-24+24}{\sqrt{(-8)^2+6^2} \times \sqrt{3^2+4^2} }\\\\\cos\theta=0\\\\\theta=90^{\circ}

Hence, this is the required solution.

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3 years ago
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You wiggle a string,that is fixed to a wall at the other end, creating a sinusoidalwave with a frequency of 2.00 Hz and an ampli
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Answer:

Explanation:

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