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A charge of -8.1 µC is traveling at a speed of 7.2 106 m/s in a region of space where there is a magnetic field. The angle between the velocity of the charge and the field is 50°. A force of magnitude 4.6 10-3 N acts on the charge. What is the magnitude of the magnetic field?

Respuesta :

Answer:

[tex]B = 1.03 \times 10^{-4} T[/tex]

Explanation:

As we know that magnetic force on a moving charge is given by the formula

[tex]F = q(\vec v \times \vec B)[/tex]

so we will have

[tex]F = qvBsin\theta[/tex]

now we will have

[tex]F = 4.6 \times 10^{-3} N[/tex]

[tex]q = 8.1 \mu C[/tex]

[tex]\theta = 50^o[/tex]

[tex]v = 7.2 \times 10^6 m/s[/tex]

now we have

[tex]4.6 \times 10^{-3} = (8.1 \times 10^{-6})(7.2 \times 10^6)Bsin50[/tex]

[tex]B = 1.03 \times 10^{-4} T[/tex]

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