A proton is propelled at 4×106 m/s perpendicular to a uniform magnetic field. 1) If it experiences a magnetic force of 4.8×10−13 N, what is the strength of the magnetic field? (Express your answer using two significant figures.)

Respuesta :

Answer:

B = 0.75 T

Explanation:

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

[tex]F = qvB[/tex]

here we have

[tex]B = \frac{F}{qv}[/tex]

here we know that

[tex]F = 4.8 \times 10^{-13} N[/tex]

[tex]q = 1.6 \times 10^{-19} C[/tex]

[tex]v = 4 \times 10^6 m/s[/tex]

now from above equation we have

[tex]B = \frac{4.8 \times 10^{-13}}{(1.6 \times 10^{-19})(4 \times 10^6)}[/tex]

[tex]B = 0.75 T[/tex]

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