Answer:
[tex]\theta=15.52^{\circ}[/tex]
Explanation:
The magnitude of the force of a moving cahrge, in our case a proton, trought a magentic field is given by:
[tex]F=|q||vB|sin(\theta)[/tex] (1)
where:
q is the proton charge ([tex]q=1.6*10^{-19} C[/tex])
v is the proton velocity ([tex]v=5*10^{5} m/s[/tex])
B is the magnetic field (B = 1.17 T)
Now, we just need to solve the equaton (1) for \theta.
[tex]\theta=sin^{-1}\left(\frac{F}{qvB}\right)[/tex]
But the force F = ma, then:
m is the mass of proton ([tex]m=1.67*10^{-27} kg[/tex])
a is the acceleration ([tex]a=1.5*10^{13} m/s^{2}[/tex])
[tex]\theta=sin^{-1}\left(\frac{m_{p}a}{q_{p}vB}\right)[/tex]
[tex]\theta=sin^{-1}\left(\frac{1.67*10^{-27}*1.5*10^{13}}{1.6*10^{-19}*5*10^{5}*1.17}\right)[/tex]
[tex]\theta=15.52^{\circ}[/tex]
I hope it helps you!
The value of the angle θ is; θ = 15.52°
The Formula for the magnitude of a force on a moving charge in a magnetic field is; F = qVB sinθ
Where;
q is charge on proton
V is velocity
B is magnetic field
θ is the angle between the velocity and the magnetic field
We are given;
B = 1.17 T
V = 500000 m/s
Acceleration; a = 1.5 × 10^(13) m/s²
Now, let's make θ the subject of the formula in the force equation.
Thus;
θ = sin^(-1) (F/qVB)
Now, we have acceleration and we know that;
F = ma
Thus;
θ = sin^(-1) (ma/qVB)
m is mass of proton = 1.67 × 10^(-27) kg
q is charge on proton = 1.6 × 10^(-19) C
Thus;
θ = sin^(-1) ((1.67 × 10^(-27) × 1.5 × 10^(13))/(1.6 × 10^(-19) × 500000 × 1.17)
θ = sin^(-1) 0.2676
θ = 15.52°
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