Answer :
Answer:
The proton was moving at a speed of 1.64 x 10⁶ m/s
Explanation:
Given;
strength of magnetic field, B = 0.280 T
circular radius, R = 6.12 cm
mass of proton, m = 1.67 x 10⁻²⁷ kg
charge of electron. q = 1.602 x 10⁻¹⁹ C
When the proton follows a circular arc, then magnetic force will be equal to the centripetal force.
Magnetic force, Fm = qvB
Centripetal force, Fr = mv²/R
Fm = Fr
qvB = mv²/R
[tex]qvB = \frac{Mv^2}{R} \\\\\frac{qBR}{M} = \frac{v^2}{v} \\\\v =\frac{qBR}{M} = \frac{1.602*10^{-19} *0.28*0.0612}{1.67*10^{-27}} =1.64 *10^6 \ m/s[/tex]
Therefore, the proton was moving at a speed of 1.64 x 10⁶ m/s
Given Information:
Magnetic field = B = 0.280 T
Radius = r = 6.12 cm = 0.0612 m
Required Information:
Speed of proton = v = ?
Answer:
Speed of proton = v = 1641.77x10³ m/s
Explanation:
Since the proton is moving in a circular path we can model it in terms of magnetic force and centripetal force.
The magnetic force is given by
F = qvB
The centripetal force is given by
F = mv²/r
equating the both equations yields,
mv²/r = qvB
mv = qBr
v = qBr/m
Where q = 1.6x10⁻¹⁹ C is the of the proton, m = 1.67x10⁻²⁷ kg is the mass of proton, B is the magnetic field and r is the radius o circular arc around which proton is moving.
v = 1.6x10⁻¹⁹*0.280*0.0612/1.67x10⁻²⁷
v = 1641772.45 m/s
v = 1641.77x10³ m/s
Therefore, the proton is moving at the speed of 1641.77x10³ m/s.