The velocity selector in in a mass spectrometer consists of a uniform magnetic field oriented at 90 degrees to a uniform electric field so that a charge particle entering the region perpendicular to both fields will experience an electric force and a magnetic force that are oppositely directed. If the uniform magnetic field has a magnitude of 11.9 ~/text{mT}11.9 mT, then calculate the magnitude of the electric field that will cause a proton entering the velocity selector at 25.325.3 km/s to be undeflected. Give your answer in units of kV/m.

Question

Jadiel Bowman

Physics

7 June, 12:17

The velocity selector in in a mass spectrometer consists of a uniform magnetic field oriented at 90 degrees to a uniform electric field so that a charge particle entering the region perpendicular to both fields will experience an electric force and a magnetic force that are oppositely directed. If the uniform magnetic field has a magnitude of 11.9 ~/text{mT}11.9 mT, then calculate the magnitude of the electric field that will cause a proton entering the velocity selector at 25.325.3 km/s to be undeflected. Give your answer in units of kV/m.

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Answers (1)

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Elaina Navarro · Answer 1

The magnitude of the electric field is 0.3011 kV/m

Explanation:

Magnetic force on the charge = qvB

where;

q is the magnitude of the charge

v is the velocity of the proton

B is the strength of the magnetic field

Electric force on the charge = Eq

where;

E is the magnitude of the electric field

q is the magnitude of the charge

Eq = qvB

E = vB

given;

v = 25.3 km/s

B = 11.9 mT

E = 25.3 x 10³ (m/s) x 11.9 x 10⁻³ (T)

E = 301.1 V/m = 0.3011 kV/m

Therefore, the magnitude of the electric field is 0.3011 kV/m