Find V of a partical in B field :O

[SoulofLoneWlf]

Homework Statement

A particle with a charge of 5.40nC is moving in a uniform magnetic field of B= -(1.20 )Tz^. The magnetic force on the particle is measured to be F=3.50×10−7N X^ + 7.60×10−7N Y^ .

Calculate the y component of the velocity of the particle.

Homework Equations

F= qv x B
F = qvBSin(theta)

The Attempt at a Solution

lost trying to use the magnitude to find also
i tried
F/qb=Vsin(theta) doesn't seem to yield results yet :/

 

[tiny-tim]

Hi SoulofLoneWlf!

A particle with a charge of 5.40nC is moving in a uniform magnetic field of B= -(1.20 )Tz^. The magnetic force on the particle is measured to be F=3.50×10−7N X^ + 7.60×10−7N Y^ .

Calculate the y component of the velocity of the particle.

Hint: call the velocity (V_x,V_y,V_z) …

then what is the x-component of the force?

 

[SoulofLoneWlf]

Hi SoulofLoneWlf!


Hint: call the velocity (V_x,V_y,V_z) …

then what is the x-component of the force?

so would it be possible for me to say v = Vx + Vy + Vz
...
Fy=qVyBsin(thetha)?

 

[tiny-tim]

Fy=qVyBsin(thetha)?

erm …

right idea , wrong result ​

 

[SoulofLoneWlf]

erm …

right idea , wrong result ​

hahaha felt that was wrong hmm let me try again i guess ><"

 

[SoulofLoneWlf]

erm …

right idea , wrong result ​

F=x^(5.4Vy*1.2Z^T)-Y^(5.4Vx*1.2TZ^)
then place value for F as 3.50x10^-7 etc and equal x to x and y to y??

 

[tiny-tim]

F=x^(5.4Vy*1.2Z^T)-Y^(5.4Vx*1.2TZ^)
then place value for F as 3.50x10^-7 etc and equal x to x and y to y??

Yup!

 

[SoulofLoneWlf]

Yup!

ur amazing thank you ^-^