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A proton, with mass 1.67 x 10-27 kg and charge +1.6 x 10-19 C, is sent with velocity 7.1 x 104 m/s in the +x direction into a region where there is a uniform electric field of magnitude 730 V/m in the +y direction. Assume that the magnetic field has no x-component and neglect gravitational effects. What is the magnitude and direction of the uniform magnetic field in the region, if the proton is to pass through undeflected?

Respuesta :

Answer:

B = 0.0102 T

Explanation:

given,                            

mass of proton =  1.67 x 10⁻²⁷ kg

charge of proton = 1.6 x 10⁻¹⁹ C

velocity = 7.1 x 10⁴ m/s in x-direction

Uniform electric field = 730 V/m in y- direction

equation of force in electric field

F = E q                          

equation of force in magnetic field

F = q V B                          

equating both force expression

E q= q V B                              

[tex]B = \dfrac{E}{V}[/tex]

[tex]B = \dfrac{730}{7.1\times 10^{4}}[/tex]

B = 0.0102 T

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