A particle with mass 1.81x10^-3 kg and a charge of 1.22x10^-8 C h...

Question

A particle with mass 1.81x10^-3 kg and a charge of 1.22x10^-8 C has, at a given instant, a velocity v=(3.00x10^4 m>s)j. What are the magnitude and direction of the particle's acceleration produced by a uniform magnetic field B=(1.63 T)i+(0.980 T)j?

Accepted Answer

Hey everyone. So this problem is working with magnetic forces. Let's see what they're giving us. And what they're asking for us, we have masks of alpha particles that's given and a given charge the alpha particle that is emitted by a radioactive substance moves into the region of a uniform magnetic field of a given magnitude and direction. So a vector field at any given moment, the alpha particle has a velocity that is also given to us. And we are asked to find the magnitude and direction of its acceleration vector at this moment where the velocity was measured. So the first thing we can do is recall that the magnitude of the force is given by Q be cross be where Q is our charge V is our velocity and B is our magnetic field. So from there, let's look at the term. So from the problem, so Q we are given its 3.20 times 10 to the minus 19 columns are speed was also given and that's 12 times 10 to the four m per second. And that is in our positive I direction positive X and then our magnetic field magnitude B is given in vector for as to Tesla's in the I direction. And then again, a positive 3.2 Tesla in the J Direction. So when we plug that in, and for each of these terms here and solve, We are left with a force of 1.2, 3 times 10 to the - newtons in the K direction. They're not asking us for the force though, they're asking us for, it's the magnitude and direction of the acceleration vector. Luckily, we can recall Newton's second law F equals M A and use that to solve for the acceleration. So the mass was also given to us in this problem As 6. times 10 to the -27 kg. And so we can rewrite this to solve for acceleration. It's going to be divided by the mass we've solved for the force vector 1.2, 3 times 10 to the -13 Newtons positive K direction. And then the mass is 6.64 times -27 kg. And that gives us 1.8, 5 times 10 to the m per second squared. And we are still in the positive K direction. So let's look at our possible answer choices. And so that aligns with choice, see we have 1.85 times 10 to the 13 m per second squared and the acceleration is in the positive Z direction. So positive K is the same as positive Z um And that's all we have for this problem. The correct answer is choice seed. All right. We'll see you in the next video.

A particle with mass 1.81x10^-3 kg and a charge of 1.22x10^-8 C has, at a given instant, a velocity v=(3.00x10^4 m>s)j. What are the magnitude and direction of the particle's acceleration produced by a uniform magnetic field B=(1.63 T)i+(0.980 T)j?

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