A proton moving at right angles to a magnetic field of 0.1 T experiences a force of 2 ｘ 10⁻¹² N. What is the speed of the proton? (Answer: 1.3 ｘ 10⁻⁸ m/s)  



Given:

Angle = θ = 90°

Magnetic Field = B = 0.1 T

Force = F = 2 ｘ 10⁻¹² N

[ ∴ Charge of proton = q = 1.6 ｘ 10⁻¹⁹ C]

To Find:

Speed of the proton = v = ?


Solution:

Since we have the magnetic force formula:

`\vec F` = q ( `\vec v` ｘ `\vec B` ) 

`\vec F` = qvB sinθ

or

v = `\frac {F}{qB sinθ}`

by putting the corresponding values

v = `\frac {2 ｘ 10⁻¹² N}{1.6 ｘ 10⁻¹⁹ C ｘ0.1 T ｘ sin90°}`

{ sin90° = 1}

v = `\frac {2 ｘ 10⁻¹² N}{0.16 ｘ 10⁻¹⁹ C T }`

v = 12.5 ｘ 10⁻¹²⁺¹⁹ m s⁻¹

v = 12.5 ｘ 10⁻⁷ m s⁻¹

v = 1.25 ｘ 10⁻⁸ m s⁻¹ ------------------ Ans.

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⭐ Henry 'H' is the unit of measurement for Inductance:

1 H = Wb A⁻¹ =  V s A⁻¹ = kg m² s⁻² A⁻²   [Wb= Weber, A=Ampere, V=Volt]


⭐ Weber 'Wb' is the unit of measurement for Magnetic Flux:

1 Wb =  V s = kg m² s⁻² A⁻¹   


⭐ Tesla 'T' is the unit of measurement for Magnetic Field:

1 T =  N m⁻¹ A⁻¹ = kg m s⁻² A⁻¹   



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