question fro dot point (1 Viewer)

[vds700]

A single positively charged particle of mass 2 x 10^-26 kg enters a 0.4 T magnetic field into the page at 3.5 x 10^5 m/s. The field covers an area of 0.3 x 0.3 m. The particle enters the field at 90 degrees, 2 cm up from the bottom left hand corner. With the aid of a diagram, determine:

a)the force on the particle as it enters the field
b)the radius of the path it takes in the field
c)The speed with which it exits the field
d)where it exits the field



Thanks in advance for any help.

 

[vds700]

does anyone know? for the first part, they havent given the charge of the particle, so u cant use F = Bqvsin(theta). I just thought there might be another way of doing it.

thanks for any help.

 

[Forbidden.]

That's why I type up the problem even if I had a scanner.
You're better off writing them here as a linear set of questions with parts like questions in a Maths textbook.

There's questions (c), (d), (e) then 1.9.2 (a) and (b), does (c), (d), (e) come after 1.9.2 or 1.9.3 ?

 

[vds700]

ok i typed the question out so theres no confusion. I appreciate any help.

 

[Forbidden.]

A single positively charged particle of mass 2 x 10^-26 kg enters a 0.4 T magnetic field into the page at 3.5 x 10^5 m/s. The field covers an area of 0.3 x 0.3 m. The particle enters the field at 90 degrees, 2 cm up from the bottom left hand corner. With the aid of a diagram, determine:

a)the force on the particle as it enters the field
b)the radius of the path it takes in the field
c)The speed with which it exits the field
d)where it exits the field



Thanks in advance for any help.

Hmmm don't have much time to solve ii but how I would solve it:

a) No actual charge ? Assume q as the positive charge of an electron. Then it's just F = qVB ...

b) r = mv/qB (derived from equating magnetic force and centripetal force)

c) and d) Use the equations of motion

 

[vds700]

Hmmm don't have much time to solve ii but how I would solve it:

a) No actual charge ? Assume q as the positive charge of an electron. Then it's just F = qVB ...

b) r = mv/qB (derived from equating magnetic force and centripetal force)

c) and d) Use the equations of motion

thanks i got a nd b out. I let q = 1.6 x 10^-19 and got the answers
a) F = qVB = 2.24 x 10^-14 N up the page.
b)r = 0.11m

However im stuck on c and d. I thought I could do c by finding the accleration of the particle using F = ma, then use v = u + at, but i dont know the time. Can someone please help. Thanks

 

[Pwnage101]

try F = Mv^2/r, as a charge in a magnetic field goes in a circle until it exits the field

 

[Forbidden.]

thanks i got a nd b out. I let q = 1.6 x 10^-19 and got the answers
a) F = qVB = 2.24 x 10^-14 N up the page.
b)r = 0.11m

However im stuck on c and d. I thought I could do c by finding the accleration of the particle using F = ma, then use v = u + at, but i dont know the time. Can someone please help. Thanks

v² - u² = -2gh ?

 

[vds700]

ok i got the answer, just needed to make mv^2/r = 2.24 x 10^-14 and sub in the info. Thanks everyone for their help.