A 1000-kg car moves west along the equator. At this location Earth's magnetic field is 3.5 × 10^−5 T

and points north parallel to Earth's surface.

 If the car carries a charge of -1.5 × 10^−3 C, how fast must it move so that the magnetic force balances 0.010% of Earth's gravitational force exerted on the car?  

units -m/s

 

mass of the car m=1000 kg

charge on the car= -1.5 × 10⁻³ C

earths magnetic field towards north B=3.5 × 10⁻⁵ T

given conditions says that the velocity and magnetic field are perpendicular to each other

thus magnetic force on car, F_mag=qvBsinθ

here θ=90

thus, F_mag=qvB

v=velocity

the magnetic force is balance by 0.01% of earths gravitational force,

F_g=GMm/r²

M=mass of earth=6.018*10²⁴ kg

G=6.67*10^-11

r=radius of earth=6.4*10⁶

thus we have, qvB=(0.01/100)*GMm/r²

thus, v=(0.01*6.67*10^-11*6.018*10²⁴*1000)/(1.5*10^-3*(6.4*10⁶)²*3.5*10^-5*100)=1.866*10⁷ m/s is the asked velocity