Respuesta :
Explanation:
It is given that,
Charge on the particle, [tex]q=5\ nC=5\times 10^{-9}\ C[/tex]
Mass of the particle, [tex]m=3\ \mu g=3\times 10^{-6}\ g=3\times 10^{-9}\ kg[/tex]
Magnetic field component, [tex]B_x=2\ mT,B_y=3\ mT,B_z=-4\ mT[/tex]
Net magnetic field, [tex]B=\sqrt{2^2+3^2+4^2}=5.38\ mT=0.00538\ T[/tex]
Speed of the particle, v = 5 km/s = 5000 m/s
Angle between velocity and magnetic field, [tex]\theta=120[/tex]
Magnetic force is given by :
[tex]F=qvB\ sin\theta[/tex]
[tex]F=5\times 10^{-9}\times 5000\ m/s\times 0.00538\times sin(120)[/tex]
[tex]F=1.16\times 10^{-7}\ N[/tex]
Acceleration of the particle is given by, [tex]a=\dfrac{F}{m}[/tex]
[tex]a=\dfrac{1.16\times 10^{-7}\ N}{3\times 10^{-9}\ kg}[/tex]
[tex]a=38.6\ m/s^2[/tex]
So, the acceleration of the particle is 38.6 m/s². Hence, this is the required solution.
The acceleration of the particle is 38.87 kg.
Net magnetic field
The net magnetic field is calculated as follows;
[tex]B_{net} = \sqrt{B_x^2 + B_y^2 + B_z^2} \\\\B_{net} = \sqrt{2^2 + 3^2 + 4^2} = 5.385 \ mT[/tex]
Magnetic force on the charge
The magnetic force on the charge is calculated as follows;
[tex]F = qvB \times sin(\theta)\\\\F = 5\times 10^{-9} \times 5\times 10^3 \times 5.385 \times 10^{-3} \times sin(120)\\\\F = 1.166 \times 10^{-7} \ N[/tex]
Acceleration of the particle
The acceleration of the particle is calculated as follows;
[tex]a = \frac{F}{m} \\\\a = \frac{1.166 \times 10^{-7}}{3 \times 10^{-9}} \\\\a = 38.87 \ kg[/tex]
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