# A proton moves in the +z direction through a uniform magnetic field that points in the -y direction and has a magnitude of 2.5 T. If the proton... moves with a speed of 5.7 x 10^6 m/s through this field, what is the STRENGTH AND DIRECTION of the force that acts on a proton.  And how would you go about finding the direction? I'm kind of confused on the right hand rule. Thanks for your help :)

The lorentz force on a charged particle is

F=q(E+vxB)

where F is the force, q is the particles charge, E is the electric field, v is the particles velocity and B is the magnetic field, whilst x denotes the cross product of the...

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The lorentz force on a charged particle is

F=q(E+vxB)

where F is the force, q is the particles charge, E is the electric field, v is the particles velocity and B is the magnetic field, whilst x denotes the cross product of the two vectors v and B.

As there is no electric field, the lorentz force equation becomes just

F=q(vxB)

The vector B, the magnetic field, is pointing in the negative y direction, thus its vector is

B=(0,-2.5T,0)

The vector v, the particle's velocity, points in the positive z direction, thus its vector is

v=(0,0,5.7x10^6 m/s)

To find the direction and magnitude of the force, you need to calculate the cross product of v and B. The way the cross product is calculated is

(vx,vy,vz)x(Bx,By,Bz)=x(vyBz-vzBy)-y(vxBz-vzBx)+z(vxBy-vyBx)

By following this calculation, you get that the only non-zero component of the force vector is

x(0x0-(5.8x10^6)x(-2.5))=(1.45x10^7)x

whilst the y and z components are both zero. This means that the vector describing the force is

F=q(1.45x10^7,0,0)=(2.32x10^-12,0,0)

Its magnitude is 2.32x10^-12N and its direction is in the positive x direction.