eg 23.2...pg 265...

when a proton has a velocity v= (2i + 3j)* 10^6 m/s it experiences a force F= -(1.28* 10^-13 k). when its velocity is along the z-axis, it experiences a force along the x-axis, what is the magnetic field?

I don't understand the how to proceed thing....how did they conclude that if magnetic force is along x- axis and velocity alongz-axis, magnetic field should be along the negative y dirn....how did they conclude its negative y dirn...im confused with these dirns so jsut help...

Velocity v along +Z axis does not feel any experience (B is on the -k direction means -Z axis)

Explanation:

When a proton has a velocity of V=(2i+3j)x10^6 m/s and also it experiences a force of F=-1.28 x 10^-13 kN then we can say that the direction of B is in the -k direction because the V makes non-zero force which is the component of j vector. Component j gives -k vector for positively charged particle. In this case it is the proton.

Now the magnitude of v is √13x10^6m/sec

F =-1.28x10^-13 kN (Given)

We know a charge of a particle is q = 1.6 x 10^-19C

From the Lorentz force equation we get,

B = F/qv = 1.28x10^-13 N /(1.6 x 10^-19 C)(3.6 x 10^3)

B = -0.222 x 10^3=-222 in -k direction

If i corresponds to the +x-axis , j to the +y-axis and k to the +z-axis then -k corresponding to -Z direction. Then velocity v along +Z axis does not feel any experience (B is on the -k direction means -Z axis)