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 :)

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The lorentz force on a charged particle is

**F**=q(**E**+**v**x**B**)

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(**v**x**B**)

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

(v*x*,v*y*,v*z*)x(B*x*,B*y*,B*z*)=**x**(v*y*B*z*-v*z*B*y*)-**y**(v*x*B*z*-v*z*B*x*)+**z**(v*x*B*y*-v*y*B*x*)

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.**

**Sources:**

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