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A 180 g model airplane charged to 18 mC and traveling at 2.2 m/s passes within 8.6 cm of a wire, nearly parallel to its path, carrying a 30 A current. What acceleration (in g's) does this interaction give the airplane?

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Answer:

[tex]a=0.2*10^{-5}g[/tex]

Explanation:

From the question we are told that:

Mass [tex]M=180=>0.18kg[/tex]

Charge [tex]Q=18mC=18*10^-^3C[/tex]

Velocity [tex]v=2.2m/s[/tex]

Length of Wire [tex]L=8.6cm=>0.086[/tex]

Current [tex]I=30A[/tex]

Generally the equation for Magnetic Field of Wire B is mathematically given by

 [tex]B=\frac{\mu_0*I}{2\pi*l}[/tex]

 [tex]B=\frac{4*3.14*10^-^7*I}{2*3.14*8.6}[/tex]

 [tex]B=6.978*10^{-5}T[/tex]

Generally the equation for Force on the plane F is mathematically given by

 [tex]F=qvB[/tex]

Therefore

 [tex]ma=qvB[/tex]

 [tex]a=\frac{qvB}{m}[/tex]

 [tex]a=\frac{18*10^{-5}83.4*6.978*10^{-5}}{0.18kg}[/tex]

 [tex]a=2.37*10^{-5}[/tex]

Therefore in Terms of g's

 [tex]a=\frac{2.37*10^{-5}}{9.8}[/tex]

 [tex]a=0.2*10^{-5}g[/tex]

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