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A charged particle is accelerated through a potential difference of 12 kV and acquires a speed of 1.0 × 106 m/s. It is then injected perpendicularly into a magnetic field of strength 0.2 T. Find the radius of the circle described by it.
- a)12 cm
- b)14 cm
- c)10 cm
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
A charged particle is accelerated through a potential difference of 12...
F(m) = qvB sin(90°)
Applied potential difference "V" = 12 kV = 12 × 103 V
Speed of a charged particle, v =1.0 × 106 m s−1
Magnetic field strength "B" = 0.2 T
As per the question, a charged particle is injected perpendicularly into the magnetic field.
We know:
qV = (½)mv2
=>m / q = (½)mv2
= (2 × 12×103)/ (1×106)2
= 24×10−9
r = mv / qB
r=(24×10−9 × 106) / 0.2
r = 12 × 10−2 m
= 12 cm
Thus, the radius of the circle is 12 cm.
Applied potential difference "V" = 12 kV = 12 × 103 V
Speed of a charged particle, v =1.0 × 106 m s−1
Magnetic field strength "B" = 0.2 T
As per the question, a charged particle is injected perpendicularly into the magnetic field.
We know:
qV = (½)mv2
=>m / q = (½)mv2
= (2 × 12×103)/ (1×106)2
= 24×10−9
r = mv / qB
r=(24×10−9 × 106) / 0.2
r = 12 × 10−2 m
= 12 cm
Thus, the radius of the circle is 12 cm.
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Community Answer
A charged particle is accelerated through a potential difference of 12...
× 10^7 m/s. What is the magnitude of the charge on the particle?
We can use the equation for the kinetic energy of a particle to solve for the charge:
K = (1/2)mv^2
where K is the kinetic energy, m is the mass of the particle, and v is its speed.
We can also use the equation for the potential difference between two points in an electric field:
V = Ed
where V is the potential difference, E is the electric field strength, and d is the distance between the two points.
Since the particle is accelerated through a potential difference of 12 kV, we can write:
K = qV
where q is the charge on the particle.
Setting these two equations equal to each other and solving for q, we get:
q = 2K/V = 2(1/2)mv^2/V
Plugging in the given values, we get:
q = 2(1/2)(mass of particle)(speed of particle)^2/(potential difference)
We don't know the mass of the particle, but we can use the fact that the kinetic energy of the particle is equal to the work done on it by the electric field:
K = qV = W
where W is the work done on the particle.
The work done on the particle is equal to the change in its potential energy:
W = ΔU = qΔV
where ΔU is the change in potential energy and ΔV is the potential difference.
Since the particle is initially at rest, its initial potential energy is zero, and its final potential energy is qΔV. Therefore:
K = qΔV
or
q = K/ΔV
Plugging in the given values, we get:
q = (1/2)(mass of particle)(speed of particle)^2/(potential difference)
We still don't know the mass of the particle, but we can cancel it out by dividing the two equations for q:
q = 2K/V = (1/2)(mass of particle)(speed of particle)^2/(potential difference)
q = 4K/(mass of particle)(speed of particle)^2
Setting these two equations equal to each other and solving for the mass of the particle, we get:
mass of particle = 8K/(q)(speed of particle)^2
Plugging in the given values, we get:
mass of particle = 8(1/2)(1.0 × 10^7)^2/(q)(12 × 10^3)
mass of particle = 2.78 × 10^-19/q
Now we can plug this expression for the mass of the particle into either of the equations for q to solve for it. Let's use the second equation:
q = (1/2)(mass of particle)(speed of particle)^2/(potential difference)
Plugging in the expression for the mass of the particle, we get:
q = (1/2)(2.78 × 10^-19/q)(1.0 × 10^7)^2/(12 × 10^3)
Solving for q, we get:
q = 3.52 × 10^-19 C
Therefore, the magnitude of the charge on the particle is 3.52 × 10^-19 C.
We can use the equation for the kinetic energy of a particle to solve for the charge:
K = (1/2)mv^2
where K is the kinetic energy, m is the mass of the particle, and v is its speed.
We can also use the equation for the potential difference between two points in an electric field:
V = Ed
where V is the potential difference, E is the electric field strength, and d is the distance between the two points.
Since the particle is accelerated through a potential difference of 12 kV, we can write:
K = qV
where q is the charge on the particle.
Setting these two equations equal to each other and solving for q, we get:
q = 2K/V = 2(1/2)mv^2/V
Plugging in the given values, we get:
q = 2(1/2)(mass of particle)(speed of particle)^2/(potential difference)
We don't know the mass of the particle, but we can use the fact that the kinetic energy of the particle is equal to the work done on it by the electric field:
K = qV = W
where W is the work done on the particle.
The work done on the particle is equal to the change in its potential energy:
W = ΔU = qΔV
where ΔU is the change in potential energy and ΔV is the potential difference.
Since the particle is initially at rest, its initial potential energy is zero, and its final potential energy is qΔV. Therefore:
K = qΔV
or
q = K/ΔV
Plugging in the given values, we get:
q = (1/2)(mass of particle)(speed of particle)^2/(potential difference)
We still don't know the mass of the particle, but we can cancel it out by dividing the two equations for q:
q = 2K/V = (1/2)(mass of particle)(speed of particle)^2/(potential difference)
q = 4K/(mass of particle)(speed of particle)^2
Setting these two equations equal to each other and solving for the mass of the particle, we get:
mass of particle = 8K/(q)(speed of particle)^2
Plugging in the given values, we get:
mass of particle = 8(1/2)(1.0 × 10^7)^2/(q)(12 × 10^3)
mass of particle = 2.78 × 10^-19/q
Now we can plug this expression for the mass of the particle into either of the equations for q to solve for it. Let's use the second equation:
q = (1/2)(mass of particle)(speed of particle)^2/(potential difference)
Plugging in the expression for the mass of the particle, we get:
q = (1/2)(2.78 × 10^-19/q)(1.0 × 10^7)^2/(12 × 10^3)
Solving for q, we get:
q = 3.52 × 10^-19 C
Therefore, the magnitude of the charge on the particle is 3.52 × 10^-19 C.
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