# Magnetic Field

1. Nov 2, 2005

### Tcat

Hi I need help finding what the acceleration is for the following problem: A particle with a mass of 1.81x10(-3) kg and a charge of 1.22 x 10(-8) C has at a given instant a velocity v=(3.11 x 10(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? I know that F=qv X B and F=ma so a= (qv X B)/m I don't know what to do from there. I think that the vector dot product is confusing me. Please help! Thank-you.

2. Nov 2, 2005

### whozum

Do you know how to take a http://mathworld.wolfram.com/CrossProduct.html" [Broken]? After that just divide by mass and multiply by charge.

Last edited by a moderator: May 2, 2017
3. Nov 2, 2005

### Tcat

u X v, but what is u and what is v in the formula? ((3.11 x 10^4)*(1.63T)) - ((3.11*10^4)*(0.980T)) = (2.02*10^4), so ((2.02*10^4)(1.22*10^8))/(1.81*10^-3) = acceleration... Is that right?

4. Nov 2, 2005

### Tcat

Could someone help me with this problem? I'm not sure what I am doing wrong?

5. Nov 2, 2005

### Gokul43201

Staff Emeritus
Look at the link in whozum's post. It teaches you how to calculate the cross-product.

The magnitude of the cross-prod, |a X b| = |a| |b| sin(Y), where Y is the angle between the vectors.

This tells you that the cross-product of parallel vectors is zero (since sin(0) = 0). And specifically, the cross-product of any vector with itself is also zero. Hence, i X i = j X j = k X k = 0. So, the second term in your calculation (which comes from the two j-components) should not be there...only the first term.