A proton moving with a velocity of `2.5 xx 10^(7) m/s`, enter a magnetic feildof 2T, making an angle of 30◦ with the magnetic field. The force acting on the proton

Explanation

Figure 29-17 shows the geometry. The first point is on the x axis, directly in front of the proton, with `theta_(1)=0^(@)`. The second point is on the y axis, with `theta_(2)=90^(@)`, and the third is in the xy plane. The magnetic field is that of a moving charged particle.
Calculate : Position 1, which is along the line of motion, has `theta_(1)=0^(@)`. Thus, `vecB_(1)=vec0`

Position 2 (at 0mm, 1mm, 0mm) is at distance `r_(2)=1mm=0.001m`. The magnetic field strength at this point is
`B=(mu_(0))/(4pi)(qvsintheta_(2))/(r_(2 ))`
`=(4pixx10^(-7)Tm//A)/(4pi)((1.60xx10^(-19)C)(1.0xx10^(7)m//s)sin90^(@))/((0.0010m)^(2))`
`=1.60xx10^(-13)T`
According to the right-hand rule, the field points in the positive z direction. Thus
`vecB_(2)=1.60xx10^(-13)hatkT`,
where `hatk` is the unit vector in the positive z direction. The field at position 3, at (1mm,1mm,0mm), also points in the z direction , but it is weaker than at position 2 both because r is larger and because q is smaller . From the geometry we know that `r_(3)=sqrt(2)mm` and `theta_(3)=45^(@)`. So
`B_(3)=0.57xx10^(-13)hatkT`
Note : These magnetic fields are indeed very small . Although Faraday asserted in 1838 that a moving charge should produce the magnetic field in its vicinity as a current-carrying wire does , it took almost 37 years to demonstrate this.