Question

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?

Answer 1

Answer:

$a=0.2*10^{-5}g$

Explanation:

From the question we are told that:

Mass $M=180=>0.18kg$

Charge $Q=18mC=18*10^-^3C$

Velocity $v=2.2m/s$

Length of Wire $L=8.6cm=>0.086$

Current $I=30A$

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

 $B=\frac{\mu_0*I}{2\pi*l}$

 $B=\frac{4*3.14*10^-^7*I}{2*3.14*8.6}$

 $B=6.978*10^{-5}T$

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

 $F=qvB$

Therefore

 $ma=qvB$

 $a=\frac{qvB}{m}$

 $a=\frac{18*10^{-5}83.4*6.978*10^{-5}}{0.18kg}$

 $a=2.37*10^{-5}$

Therefore in Terms of g's

 $a=\frac{2.37*10^{-5}}{9.8}$

 $a=0.2*10^{-5}g$