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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.

Respuesta :

onyebuchinnaji

Answer:

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

abidemiokin

The magnitude of the electric field that will cause a proton entering the velocity selector at 25.325.3 km/s to be undeflected is 0.3011kV/m

The formula for calculating the magnetic force on the charge is expressed as:

[tex]M=qvB[/tex] where:

 q is the magnitude of the charge

v is the velocity of the proton

B is the magnetic field strength

The formula for the electric force on the charge is expressed as:

[tex]F=qE[/tex]

q is the charge

Equating both expressions

[tex]qE=qvB\\E=vB[/tex]

E is the required magnitude of the electric field.

Given the following parameters

[tex]v=25.325.3 km/s \\B=11.9 \times 10^{-3}T[/tex]

Substitute the given parameters into the formula to have:

[tex]E = 25.325.3 \times 0.0119\\E= 0.3011 kV/m[/tex]

Hence the magnitude of the electric field that will cause a proton entering the velocity selector at 25.325.3 km/s to be undeflected is 0.3011kV/m

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