You have a charged particle moving through perpendicular electric and magnetic f

Question

You have a charged particle moving through perpendicular electric and magnetic fields. If a is odd, then the particle is an ELECTRON.  If a is even, then the particle is a proton.  At the begining, the particle is moving North at a speed of (m) km/s through a magnetic field that has a magnitude of (b/100) T.  If b is odd, then the magnetic field points DOWN.  If b is even, then the magnetic field points up.  

Note: you may completley ignore gravity in this problem.

a = 5

b = 7

m = 25

n = 85

a) What is the direction of the magnetic force on the charge?

b) What is the magnitude of the magnetic force on the charge?

c) What electric field (magnetic and direction) would be required to cancel out the magnetic force on the charge (resulting in a net force of zero)?

Answer's must be clear and explained to the best of your ability.

Thanks

Explanation / Answer

a) is odd, then the particle is an electron

if (b) is even, then the magnetic field points UP;

and f = q(v cross b)

so direction = westward ie along -x

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.