Question

a 2.99 x 10^-6 C charge is moving perpendicular (90) to the earth magnetic field (5.00 x 10^-5 T). If the force on it is 2.14 x 10^-8 N, how far is it moving
? m/s

Answer 1

Hi! So do you know the formula we should be using?

Answer 2

no

Answer 3

F = qvB sin\(\theta\)

Have you seen this formula before?

q is the charge
v is the velocity
B is the magnetic field
\(\theta\) is the angle between the velocity and magnetic field

Answer 4

YES

Answer 5

PLUG IT IN FOR ME

Answer 6

AND THEN ILL ANSWER IT

Answer 7

okay okay I'll do it for you this one time, but there's a reason why I haven't been

Plugging in is the easy part and you should be able to do it yourself. Figuring out the equation requires practice and all

But if you can't plug it in, or figure out what to plug in, then it'll be challenging as you continue your physics, chemistry, math classes

But here we go

The question says the charge = q = 2.99*10^-6 C
Magnetic field = B = 5.00 * 10^-5 T
\(\theta = 90^o\)
And we're solving for velocity or v

Do you follow so far?

Answer 8

I GUESS

Answer 9

just answer this one for me and ill figure the next one

Answer 10

So plug and chug
Forgot the force but it's also given in the question which is 2.14*10^-8 N

F = qvB sin\(\theta\)

2.14*10^-8 = (2.99*10^-6) * v * (5.00*10^-5) sin\((90^o)\)

So solve for v.

Hint: sin 90 = 1

Answer 11

Let me know what you get
First multiply the numbers on the right side of the equal sign

And then divide by that number on both sides so that way you can get 'v' all by itself

Answer 12

143

Answer 13

That's right!