91Ó°ÊÓ

$$ B ≈ 1.83 × 10^{-5} \,\mathrm{T} $$ #tag_title#Step 2: Calculate the magnetic force acting on the airplane#tag_content#Next, we will calculate the magnetic force (F) acting on the airplane using the Lorentz force formula: $$ F = qvB \sin{\theta} $$ where \(F\) is the magnetic force, \(q\) is the electric charge of the airplane (1.5 µC = 1.5 × 10^{-6} C), \(v\) is the velocity of the airplane (65 m/s), \(B\) is the magnetic field strength (1.83 × 10^{-5} T), and \(\theta\) is the angle between the velocity and the magnetic field (90°). Since the angle is 90°, the sine value is 1, and the formula simplifies as: $$ F = qvB $$ Now, we can plug in the values and solve for \(F\): $$ F = (1.5 × 10^{-6}\,\mathrm{C})(65\,\mathrm{m/s})(1.83 × 10^{-5}\,\mathrm{T}) $$ $$ F ≈ 1.78 × 10^{-9}\,\mathrm{N} $$ #tag_title#Step 3: Determine the acceleration using Newton's second law#tag_content#Finally, we will determine the acceleration (a) of the toy airplane using Newton's second law of motion: $$ F = ma $$ where \(F\) is the magnetic force (1.78 × 10^{-9} N), \(m\) is the mass of the airplane (0.38 kg), and \(a\) is the acceleration. We need to solve for \(a\). Rearrange the formula to solve for \(a\): $$ a = \frac{F}{m} $$ Now, plug in the values and solve for \(a\): $$ a = \frac{1.78 × 10^{-9}\,\mathrm{N}}{0.38\,\mathrm{kg}} $$ $$ a ≈ 4.68 × 10^{-9}\,\mathrm{m/s^2} $$ #Answer# The acceleration experienced by the toy airplane due to the magnetic field produced by the wire is approximately \(4.68 × 10^{-9}\,\mathrm{m/s^2}\).