Question

A charged particle moving through a magnetic field at right angles to the field with a speed of 25.7 m/s experiences a magnetic force of 2.98 10-4 N. Determine the magnetic force on an identical particle when it travels through the same magnetic field with a speed of 4.64 m/s at an angle of 29.2° relative to the magnetic field.

Answer 1

The magnetic force would be:

`F\approx 2.625\,\,10^(-5)\,\,N`

Step-by-step explanation:

Recall that the magnetic force on a charged particle (of charge q) moving with velocity (v) in a magnetic field B, is given by the vector product:

F = q v x B

(where the bold represents vectors)

the vector product involves the sine of the angle (
`\theta`) between the vectors, so we can write the relationship between the magnitudes of these quantities as:

`F=q\,v\,B\,sin(\theta)`

Therefore replacing the known quantities for the first case:

`F=q\,v\,B\,sin(\theta)\\2.98\,\,10^(-4) \,\,N=q\,(25.7\,\,m/s)\,B\,sin(90^o)\\2.98\,\,10^(-4) \,\,N=q\,(25.7\,\,m/s)\,B\\q\,\,B=(2.98\,\,10^(-4) )/(25.7) \,(N\,\,s)/(m)`

Now, for the second case, we can find the force by using this expression for the product of the particle's charge times the magnetic field, and the new velocity and angle:

`F=q\,v\,B\,sin(\theta)\\F=q\,(4.64\,\,m/s)\,B\,sin(29.2^o)\\F=q\,B(4.64\,\,m/s)\,\,sin(29.2^o)\\F=(2.98\,\,10^(-4) )/(25.7) \,(4.64\,\,m/s)\,\,sin(29.2^o)\\F\approx 2.625\,\,10^(-5)\,\,N`