The vertical component of the magnetic induction in the earth's magnetic field at hobart is approximately 6×10-5t upward. what electric field is set up in a car travelling on a level surface at 100 km h-1due to this magnetic field? which side or end of the car is positively charged? approximately what p. d. is created across a car of typical size?

Explanation:

Magnetic field B = 6 X 10⁻⁵ T.

Width of car = L (Let )

Velocity of car v  = 100 km/h

= 27.78 m /s

induced emf across the body ( width )  of the car

= BLv

= 6 X 10⁻⁵ L X 27.78

166.68 X 10⁻⁵ L

Induced electric field across the width

= emf induced / L

E =  166.68 X 10⁻⁵ N/C

We suppose breadth of a typical car = 1.5 m

potential difference induced

= 166.68 x 1.5 x 10⁻⁵

250 x 10⁻⁵ V

= 2.5 milli volt.

The side of the car which is positively charged depends on the direction in which car is moving , whether it is moving towards the north or south.

answer: the correct answer id option(d).

explanation:

snell's law refraction:

the angle of refraction =

the angle of incidence =

refractive index of air = = 1

refractive index of glycerol = 1.47

from the given options, the most closest answer to 63.72° is 65° of option (d).

speed

ratio of distance traveled to time interval

net force

vector sum of forces on object

object

source of diverging light rays either luminous or illuminated

don't remove my answer im trying my best to answer them

explanation: