A particle with a mass of 6.64x10^-27 kg and a charge of +3.2x10^-19 C is accelerated from rest through a potential difference of 2.45x10^6 V. the particle then enters a uniform magnetic field of strength 1.60 T.
a) Find the particle's gain in KE as it moves through the potential difference
b) What is the velocity of the particle as it enters the magnetic field?
c) If the particle's velocity is perpendicular to the magnetic field at all times, what is the magnitude of the magnetic force exerted on the particle as it passes through the magnetic field.
KE=mv²/2= qU
v=sqrt(2qU/m)=…
F=qvB=…
a) Well, the particle is really getting charged up with all these numbers! To find the gain in kinetic energy (KE), we can use the formula:
KE = qΔV
Where q is the charge of the particle and ΔV is the potential difference. Plugging in the given values:
KE = (+3.2x10^-19 C) * (2.45x10^6 V)
Now my math skills are feeling electrified. Let me calculate that for you:
KE = (3.2 * 2.45) * (10^-19 * 10^6) C * V
= 7.84 * 10^-13 J
So, the particle's gain in KE is approximately 7.84x10^-13 joules. It's getting quite the boost!
b) To find the velocity as the particle enters the magnetic field, we can use the formula:
qΔV = 1/2 mv^2
Where q is the charge of the particle, ΔV is the potential difference, m is the mass of the particle, and v is its velocity. Rearranging the formula:
v = √((2qΔV) / m)
Plugging in the values:
v = √((2 * 3.2x10^-19 C * 2.45x10^6 V) / 6.64x10^-27 kg)
Hold on to your electrons, let me calculate that for you:
v = √((6.4 * 2.45) * (10^-19 * 10^6) C * V / (6.64 * 10^-27) kg)
= √(15.68 * 10^-13 C * V / 6.64 * 10^-27 kg)
≈ √(2.36 * 10^14 C * V / 10^-13 kg)
≈ √(2.36 * 10^27 m^2 * s^-2 / kg)
Wow, that's a mouthful! But let's not go nuts. I'll simplify it for you:
v ≈ 1.54 * 10^13 m/s
So, the velocity of the particle as it enters the magnetic field is around 1.54x10^13 meters per second. It's zipping around faster than a clown car!
c) Since the particle's velocity is perpendicular to the magnetic field, we can use the formula for the magnetic force:
F = qvB
Where q is the charge of the particle, v is its velocity, and B is the magnetic field strength. Let's plug in the values:
F = (+3.2x10^-19 C) * (1.54x10^13 m/s) * (1.6 T)
Let me do some juggling with these numbers:
F = (3.2 * 1.54 * 1.6) * (10^-19 * 10^13) C * m * s^-1 * T
= (7.84 * 10^-6) * (10^-6) C * m * s^-1 * T
= 7.84x10^-12 N
So, the magnitude of the magnetic force exerted on the particle as it passes through the magnetic field is approximately 7.84x10^-12 newtons. It's quite a shocking force!
Keep in mind that my calculations are just for fun and to bring a smile to your face. Always double-check with reliable sources!
a) To find the particle's gain in kinetic energy (KE), we can use the formula:
ΔKE = qΔV
where ΔV is the potential difference and q is the charge of the particle.
Given:
ΔV = 2.45x10^6 V
q = +3.2x10^-19 C
Substituting the values into the formula:
ΔKE = (3.2x10^-19 C) * (2.45x10^6 V)
Calculating the value:
ΔKE = 7.84x10^-13 J
Therefore, the particle's gain in kinetic energy as it moves through the potential difference is 7.84x10^-13 J.
b) To find the velocity of the particle as it enters the magnetic field, we can use the equation for the kinetic energy of a particle:
KE = 1/2 mv^2
We can solve this equation for v:
v = √(2KE / m)
Given:
KE = 7.84x10^-13 J (obtained in part a)
m = 6.64x10^-27 kg
Substituting the values into the equation:
v = √[(2 * 7.84x10^-13 J) / (6.64x10^-27 kg)]
Calculating the value:
v ≈ 1.56x10^4 m/s
Therefore, the velocity of the particle as it enters the magnetic field is approximately 1.56x10^4 m/s.
c) If the particle's velocity is perpendicular to the magnetic field at all times, the magnitude of the magnetic force exerted on the particle can be found using the equation:
F = qvB
where q is the charge of the particle, v is its velocity, and B is the magnetic field strength.
Given:
q = +3.2x10^-19 C
v = 1.56x10^4 m/s (obtained in part b)
B = 1.60 T
Substituting the values into the equation:
F = (3.2x10^-19 C) * (1.56x10^4 m/s) * (1.60 T)
Calculating the value:
F ≈ 8.19x10^-14 N
Therefore, the magnitude of the magnetic force exerted on the particle as it passes through the magnetic field is approximately 8.19x10^-14 N.
To answer these questions, we need to apply the principles of energy conservation and the Lorentz force.
a) To find the particle's gain in kinetic energy (KE), we can use the equation:
Gain in KE = Charge × Potential Difference
Given:
Mass of the particle (m) = 6.64 × 10^-27 kg
Charge of the particle (Q) = +3.2 × 10^-19 C
Potential Difference (V) = 2.45 × 10^6 V
Substituting the values into the equation:
Gain in KE = Q × V
= (+3.2 × 10^-19 C) × (2.45 × 10^6 V)
≈ 7.84 × 10^-13 J
Therefore, the particle's gain in kinetic energy as it moves through the potential difference is approximately 7.84 × 10^-13 J.
b) The velocity of the particle as it enters the magnetic field can be found using the equation:
Force = Charge × Velocity × Magnetic Field Strength
However, since the particle is moving through the potential difference, it gains kinetic energy and its velocity will not be zero. We can use the gained kinetic energy to find the velocity.
The gained kinetic energy can be found using the equation:
Gain in KE = 0.5 × Mass × Velocity^2
Given:
Mass of the particle (m) = 6.64 × 10^-27 kg
Gain in KE = 7.84 × 10^-13 J
Substituting these values into the equation and solving for velocity:
7.84 × 10^-13 J = 0.5 × (6.64 × 10^-27 kg) × v^2
v^2 = (2 × 7.84 × 10^-13 J) / (6.64 × 10^-27 kg)
v^2 ≈ 2.365 × 10^14 m^2/s^2
v ≈ √(2.365 × 10^14) m/s
v ≈ 1.537 × 10^7 m/s
Therefore, the velocity of the particle as it enters the magnetic field is approximately 1.537 × 10^7 m/s.
c) The magnetic force exerted on a charged particle moving perpendicular to a magnetic field is given by the equation:
Force = Charge × Velocity × Magnetic Field Strength
Given:
Charge of the particle (Q) = +3.2 × 10^-19 C
Velocity of the particle (v) = 1.537 × 10^7 m/s
Magnetic Field Strength (B) = 1.60 T
Substituting the values into the equation:
Force = (3.2 × 10^-19 C) × (1.537 × 10^7 m/s) × (1.60 T)
≈ 7.81 × 10^-12 N
Therefore, the magnitude of the magnetic force exerted on the particle as it passes through the magnetic field is approximately 7.81 × 10^-12 N.