Physics Question (1 Viewer)

theodore0307 · Dec 18, 2013

Can someone help me with this question

anomalousdecay · Dec 18, 2013

The 10A wire exerts a force moving the 3A wire to the left. The 2A wire exerts a force moving the 3A wire to the right.

So using the formula:

$\frac{F}{l}=\frac{kI_1I_2}{d}$

You get the force exerted by each wire.

Then you need to use vector components to analyse the direction the wire moves in and the net force.

Does this help?

If you need I can do the question with you step-by-step.

theodore0307 · Dec 18, 2013

anomalousdecay said:
The 10A wire exerts a force moving the 3A wire to the left. The 2A wire exerts a force moving the 3A wire to the right.

So using the formula:

$\frac{F}{l}=\frac{kI_1I_2}{d}$

You get the force exerted by each wire.

Then you need to use vector components to analyse the direction the wire moves in and the net force.

Does this help?

If you need I can do the question with you step-by-step.

Can you please. Thanks

Sent from my GT-I9100 using Tapatalk

anomalousdecay · Dec 19, 2013

When moving to the right we get:

$\frac{F}{l}= (2x10^{-7}N$ $ A^{-2})(2A)(3A)/(0.01m) $to the right $\\ \\ \\ F= 1.2 x 10^{-4} N $to the right$$

To the left we get:

$\frac{F}{l}=(2x10^{-7}N$ $ A^{-2})(10A)(3A)/{(0.02m)} $to the left$ \\ \\ \\ F= 3 x 10^{-4} N $to the left $$

So the Net Force is:

$\Sigma F= 3x 10^{-4}N $to the left$ - 1.2x10^{-4}N $to the left$ \\ \\ \\ \Sigma F=1.8x10^{-4}N $ to the left$$

$The net force on the middle wire is $1.8 x 10^{-4} N$ to the left$

Physics Question (1 Viewer)

theodore0307

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anomalousdecay

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theodore0307

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anomalousdecay

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