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alv

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Member advised to use the homework template for posts in the homework sections of PF.

I am a teacher and I can't work this one out. I am trying to answer (a). I have only included the further parts of the question (b, c and d) so you can see why some parts of the question are included.

A single positively charged particle, mass 4.6 x 10

(a) Calculate the force on the particle while it is in the field.

(b) Calculate the radius of the path the particle takes in the field.

(c) Calculate the speed with which the particle exits the field.

(d) Where will the particle exit the field?

There is no indication as to what the positive charged particle is. It's too massive for a proton. So I don't know what the charge on it is. I can't see how would be meant to assume that it has the charge of a proton therefore I don't know what the charge is.

I have tried to use F = mv2/r but I don't know the radius.

I could use F = qvB sinθ but as I said, I don't know the charge

I have calculated E from v = E/B so E = vB = 2.5 x 10^6 x 4 = 1 x 10^7 but then I don't know what to do with it.

There is F = BIl sinθ but I don't know what the current is (or length for that matter).

The answer is 1.8 x 10

Your help would be greatly appreciated.

A single positively charged particle, mass 4.6 x 10

^{27}kg, enters a 4.0 T magnetic field into the page at 2.5 x 10^{6}m/s. The field covers an area of 0.1 m x 0.1 m. It enters at 90 deg, 0.05 m up from the bottom left corner.(a) Calculate the force on the particle while it is in the field.

(b) Calculate the radius of the path the particle takes in the field.

(c) Calculate the speed with which the particle exits the field.

(d) Where will the particle exit the field?

There is no indication as to what the positive charged particle is. It's too massive for a proton. So I don't know what the charge on it is. I can't see how would be meant to assume that it has the charge of a proton therefore I don't know what the charge is.

I have tried to use F = mv2/r but I don't know the radius.

I could use F = qvB sinθ but as I said, I don't know the charge

I have calculated E from v = E/B so E = vB = 2.5 x 10^6 x 4 = 1 x 10^7 but then I don't know what to do with it.

There is F = BIl sinθ but I don't know what the current is (or length for that matter).

The answer is 1.8 x 10

^{-12}but I don't know how to get it and need to so I can show my students.Your help would be greatly appreciated.