Can anyone help with this magnetism problem?

[kbraith3]

A 2.53*10^(-6) C charged particle with kinetic energy of 0.0929 J is fired into a uniform magnetic field of magnitude 0.147 T. If the particle moves in a circular path of radius 3.38 m, determine its mass in kg. (answer should be within 2*10^-14 kg)

The formula r=mv/qb comes to mind where r is radius (in m), m is mass (in kg), v is velocity (in m/s), q is charge in coulombs, and b is the magnetic field in teslas. Another important formula is Kinetic energy is 0.5mv^(2). where m is mass in kg and v is is m/s (I think).

I tried to solve this but I run into the problem of not being able to isolate the m to solve for it. I know that 0.0929 J = 0.5mv^(2) but I still have two unknowns and can't solve for the mass. Otherwise I could insert into r=mv/qb, such as 3.38m=mv/(2.53*10^(-6) C)(0.147 T) It seems like I would need to be given the velocity to be able to solve for the mass, can anyone help??

 

[Merlin3189]

Hello and welcome to PF. Hope you find it interesting and helpful.

I tried to solve this but I run into the problem of not being able to isolate the m to solve for it. I know that 0.0929 J = 0.5mv^(2) but I still have two unknowns and can't solve for the mass.

Well, if you have two unknowns, you need two equations - and you have them.
Maybe if you tried some algebraic manipulation, so that you could substitute one into the other, eliminating one variable, you would be left with one equation and one unknown.

 

[kbraith3]

Hello and welcome to PF. Hope you find it interesting and helpful.Well, if you have two unknowns, you need two equations - and you have them.
Maybe if you tried some algebraic manipulation, so that you could substitute one into the other, eliminating one variable, you would be left with one equation and one unknown.

I think I see what you're saying, but I'm not sure how to do it. The V in one equation and the V^2 in the other seems to throw everything else off...

 

[Merlin3189]

If you got something like $x=y^2 +3$ how would you get what y= ?

 

[kbraith3]

If you got something like $x=y^2 +3$ how would you get what y= ?

if x=y^(2)+3 if would just subtract three from both sides yielding x-3=y^(2), I would then square root both sides to get Sqrt(x-3)=y

 

[Merlin3189]

Exactly. So do likewise with $K= \frac 1 2 mv^2$

 

[kbraith3]

Exactly. So do likewise with $K= \frac 1 2 mv^2$

nevermind I figured it out. You can just take both equations and isolate for m on both of them. When you do this you realize that on one side rqb/v=m and on the other side 2Ke/v^(2)=m, therefore rqb/v=2Ke/v^(2). Furthering the work, you can multiply rqb/v by v^(2) on both sides to eliminate it from the denominator on the 2Ke/v^(2) leaving rqbv=2Ke. From here we can isolate the v, v=2Ke/rqb. At this point we have all the variables and can solve for v. Once we have v we can substitute it back into our original equation to solve for m.

 

[kbraith3]

it turns out v=147,805.7 m/s. Substituting back in we get 3.38 m = m(147,805.7 m/s)/((2.53E-6 C)(0.147 T)). m=8.505*10^-12 kg.

 

[Merlin3189]

Fine. I haven't put the numbers in yet - just done the algebra. I'll get my specs & a calculator and check.

yep. I get the same.