2005 Q21b) (1 Viewer)

JaxsenW · Feb 16, 2018

Hey Guys, just wondering if someone could give me a straight value answer for this question?

I have done the working and am confident with my answer but have no way of checking the answer (I cant find any solutions for 2005)

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captainhelium · Feb 16, 2018

I got 221.35 A

I guess this was my working. I don't know if you need it but I'll just put it here

$Use Ampere's Force Law$

$F = \frac {kI_1I_2l}{d}$

$However, both wires are connected to the same car battery so the currents should be the same. Therefore, we can rewrite the equation as$

$F = \frac{kI^2l}{d}$

$Since the question is asking for minimum current for one tube to jump, it must overcome weight force and so we can just substitute F with mg$

$mg = \frac{kI^2l}{d}$

$Rearranging gives$

$I^2 = \frac {(mg)(d)}{kl}$

$Just sub the values in and take the square root and you should get I = 221.35 A$

JaxsenW · Feb 17, 2018

captainhelium said:
I got 221.35 A

I guess this was my working. I don't know if you need it but I'll just put it here

$Use Ampere's Force Law$

$F = \frac {kI_1I_2l}{d}$

$However, both wires are connected to the same car battery so the currents should be the same. Therefore, we can rewrite the equation as$

$F = \frac{kI^2l}{d}$

$Since the question is asking for minimum current for one tube to jump, it must overcome weight force and so we can just substitute F with mg$

$mg = \frac{kI^2l}{d}$

$Rearranging gives$

$I^2 = \frac {(mg)(d)}{kl}$

$Just sub the values in and take the square root and you should get I = 221.35 A$

Thankyou, this is what I got

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2005 Q21b) (1 Viewer)

JaxsenW

Active Member

captainhelium

water enthusiast

JaxsenW

Active Member

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