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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?

Respuesta :

Answer:

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.

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