Anonymous
more than 1 week ago
A proton accelerated through a potential difference of 3.0kV enters a magnetic field of flux density 0.1T, in a plane perpendicular to the motion of the proton. Calculate?
a) The speed of the proton on entering the magnetic field?
b) The radius of the circular path that the particle describe in the magnetic field?
b) The radius of the circular path that the particle describe in the magnetic field?
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ChatGPT
more than 1 week ago
The force experienced by a charged particle moving through a magnetic field is given by the equation:
F = qvB
Where:
F = force
q = charge of the particle (in this case, the charge of a proton is 1.6 x 10^-19 C)
v = velocity of the particle
B = magnetic flux density
The velocity of the proton can be calculated using the equation for kinetic energy:
KE = qV
Where:
KE = kinetic energy
q = charge of the particle
V = potential difference
Rearranging the equation for velocity:
v = sqrt(2KE/m)
Where:
m = mass of the proton (1.67 x 10^-27 kg)
Substitute the values into the equations:
KE = 1.6 x 10^-19 C * 3.0 x 10^3 V = 4.8 x 10^-16 J
v = sqrt(2 * 4.8 x 10^-16 J / 1.67 x 10^-27 kg) = 1.6 x 10^7 m/s
Now, calculate the force experienced by the proton:
F = 1.6 x 10^-19 C * 1.6 x 10^7 m/s * 0.1 T = 1.6 x 10^-12 N
Therefore, the force experienced by the proton is 1.6 x 10^-12 N.
F = qvB
Where:
F = force
q = charge of the particle (in this case, the charge of a proton is 1.6 x 10^-19 C)
v = velocity of the particle
B = magnetic flux density
The velocity of the proton can be calculated using the equation for kinetic energy:
KE = qV
Where:
KE = kinetic energy
q = charge of the particle
V = potential difference
Rearranging the equation for velocity:
v = sqrt(2KE/m)
Where:
m = mass of the proton (1.67 x 10^-27 kg)
Substitute the values into the equations:
KE = 1.6 x 10^-19 C * 3.0 x 10^3 V = 4.8 x 10^-16 J
v = sqrt(2 * 4.8 x 10^-16 J / 1.67 x 10^-27 kg) = 1.6 x 10^7 m/s
Now, calculate the force experienced by the proton:
F = 1.6 x 10^-19 C * 1.6 x 10^7 m/s * 0.1 T = 1.6 x 10^-12 N
Therefore, the force experienced by the proton is 1.6 x 10^-12 N.
1
Google Gemini AI
more than 1 week ago
1. Proton's kinetic energy:
Kinetic energy = qV = (1.6 x 10^-19 C)(3000 V) = 4.8 x 10^-16 J
2. Proton's velocity:
Kinetic energy = 1/2 - mv^2
4.8 x 10^-16 J = 1/2 - (1.67 x 10^-27 kg) - v^2
v = 1.2 x 10^6 m/s
3. Magnetic force on proton:
Magnetic force = qvB
Magnetic force = (1.6 x 10^-19 C)(1.2 x 10^6 m/s)(0.1 T)
Magnetic force = 1.92 x 10^-13 N
4. Radius of circular path:
Magnetic force = mv^2/r
1.92 x 10^-13 N = (1.67 x 10^-27 kg)(1.2 x 10^6 m/s)^2 / r
r = 0.01 m = 1 cm
Therefore, the proton will move in a circular path with a radius of 1 cm.
Kinetic energy = qV = (1.6 x 10^-19 C)(3000 V) = 4.8 x 10^-16 J
2. Proton's velocity:
Kinetic energy = 1/2 - mv^2
4.8 x 10^-16 J = 1/2 - (1.67 x 10^-27 kg) - v^2
v = 1.2 x 10^6 m/s
3. Magnetic force on proton:
Magnetic force = qvB
Magnetic force = (1.6 x 10^-19 C)(1.2 x 10^6 m/s)(0.1 T)
Magnetic force = 1.92 x 10^-13 N
4. Radius of circular path:
Magnetic force = mv^2/r
1.92 x 10^-13 N = (1.67 x 10^-27 kg)(1.2 x 10^6 m/s)^2 / r
r = 0.01 m = 1 cm
Therefore, the proton will move in a circular path with a radius of 1 cm.
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