Answer:
The magnetic force would be:
[tex]F\approx 2.625\,\,10^{-5}\,\,N[/tex]
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 ([tex]\theta[/tex]) between the vectors, so we can write the relationship between the magnitudes of these quantities as:
[tex]F=q\,v\,B\,sin(\theta)[/tex]
Therefore replacing the known quantities for the first case:
[tex]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=\frac{2.98\,\,10^{-4} }{25.7} \,\frac{N\,\,s}{m}[/tex]
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:
[tex]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=\frac{2.98\,\,10^{-4} }{25.7} \,(4.64\,\,m/s)\,\,sin(29.2^o)\\F\approx 2.625\,\,10^{-5}\,\,N[/tex]
