Answers
Explanation:
Charge on proton, q₁ = e
Charge on alpha particles, q₂ = 2e
The magnetic force is given by :
Here,
For proton, ..........(1)
For alpha particle, ..........(2)
Since, a proton (charge e), traveling perpendicular to a magnetic field, experiences the same force as an alpha particle. So,
So, the ratio of the speed of proton to the alpha particle is 2 : 1 .Hence, this is the required solution.
Final answer:
If a proton and an alpha particle experience the same force in a magnetic field, the proton must be traveling at twice the speed of the alpha particle. This is because the force exerted by a magnetic field on a moving charge depends on the charge of the particle, the speed of the particle, and the strength of the magnetic field.
Explanation:
The force exerted by a magnetic field on a moving charge depends on the charge of the particle, the speed of the particle, and the strength of the magnetic field. Given that a proton (charge e) and alpha particle (charge 2e) experience the same force in the same magnetic field, we can create an equation to solve for their speed ratio.
The force on a particle due to a magnetic field is given by F = qvB where q is the charge, v is the speed, and B is the magnetic field. Since the force on the proton and alpha particle are the same, we can set their force equations equal to each other.
This means that e * v_proton * B = 2e * v_alpha * B. Simplifying, the ratio v_proton/v_alpha = 2.
Therefore, the proton is moving twice as fast as the alpha particle.
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Related Questions
wo parallel plates of area 100cm2are given charges of equal magnitudes 8.9 ×10−7C but opposite signs. The electric field within the dielectric material filling the space between the plates is 1.4 ×106V/m. (a) Calculate the dielectric constant of the material. (b) Determine the magnitude of the charge induced on each dielectric surface.
A 0.010 kg ball is shot from theplunger of a pinball machine.Because of a centripetal force of0.025 N, the ball follows acirculararc whose radius is 0.29 m. What isthe speed of theball?
) An electron moving along the x-axis enters a magnetic field. If the electron experiences a magnetic deflection in the -y direction, what is the direction of the magnetic field in this region
A parallel-plate capacitor is charged and then disconnected from the battery. By what factor does the stored energy change when the plate separation is then doubled?
(b) F/4,
(c) F,
(d) 2F,
(e) 4F,
(f) None of the above.
Answers
Answer:
F/2
Explanation:
In the first case, the two charges are Q1 and Q2 and the distance between them is r. K is the Coulomb's constant
Hence;
F= KQ1Q2/r^2 ------(1)
Where the charge on Q1 is doubled and the distance separating the charges is also doubled;
F= K2Q1 Q2/(2r)^2
F2= 2KQ1Q2/4r^2 ----(2)
F2= F/2
Comparing (1) and (2)
The magnitude of force acting on each of the two particles is;
F= F/2
Answers
Answer:
r₁/r₂ = 1/2 = 0.5
Explanation:
The resistance of a wire is given by the following formula:
R = ρL/A
where,
R = Resistance of wire
ρ = resistivity of the material of wire
L = Length of wire
A = Cross-sectional area of wire = πr²
r = radius of wire
Therefore,
R = ρL/πr²
FOR WIRE A:
R₁ = ρ₁L₁/πr₁² -------- equation 1
FOR WIRE B:
R₂ = ρ₂L₂/πr₂² -------- equation 2
It is given that resistance of wire A is four times greater than the resistance of wire B.
R₁ = 4 R₂
using values from equation 1 and equation 2:
ρ₁L₁/πr₁² = 4ρ₂L₂/πr₂²
since, the material and length of both wires are same.
ρ₁ = ρ₂ = ρ
L₁ = L₂ = L
Therefore,
ρL/πr₁² = 4ρL/πr₂²
1/r₁² = 4/r₂²
r₁²/r₂² = 1/4
taking square root on both sides:
r₁/r₂ = 1/2 = 0.5
Final answer:
The ratio of the radius of wire A to the radius of wire B is 1/2.
Explanation:
The resistance of a wire is given by the formula R = ρl/A, where R is resistance, ρ is resistivity, l is length, and A is the cross-sectional area of the wire. When the wire has a circular cross-section, the area can be calculated by the formula A = πr². The resistance of the wire then becomes: R = ρl/(πr²). If the resistance of wire A is four times that of wire B, we can set up the equation 4RB = RA. Substituting the expression for resistance, we get 4(ρl/(πrB²)) = ρl/(πrA²). Simplifying, we find that the ratio of the radius of wire A to the radius of wire B is one-half, or rA/rB = 1/2.
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Answers
Answer:
we see it is a linear relationship.
Explanation:
The magnetic flux is u solenoid is
B = μ₀ N/L I
where N is the number of loops, L the length and I the current
By applying this expression to our case we have that the current is the same in all cases and we can assume the constant length. Consequently we see that the magnitude of the magnetic field decreases with the number of loops
B = (μ₀ I / L) N
the amount between paracentesis constant, in the case of 4 loop the field is worth
B = cte 4
N B
4 4 cte
3 3 cte
2 2 cte
1 1 cte
as we see it is a linear relationship.
In addition, this effect for such a small number of turns the direction of the field that is parallel to the normal of the lines will oscillate,
person with male and female traits?
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Answers
Answers
Answer: With no friction, the box will accelerate down the ramp
Explanation:
Answers
Answer:
Emf induced in the loop is 0.02V
Explanation:
To get the emf of induced loop, we have to use faraday's law
ε = - dΦ/dt
To get the flux, we use;
Φ = BA cos(θ)
B = The uniform magnetic field
A = Area of rectangular loop
θ = angle between magnetic field and normal to the plane of loop
substitute the flux equation (Φ) into the faraday's equation
we have ε = - d(BA cos(θ)) / dt
ε = BA sinθ dθ/dt
from the question;B = 0.18T, A=0.15m2, θ = π/2 ,dθ/dt = 0.75rad/s
Our equation will now look like this;
ε = (0.18T) (0.15m2) (sin(π/2)) (0.75rad/s)
ε = 0.02V