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Afina-wow [57]
3 years ago
7

A proton moving at 6.60 106 m/s through a magnetic field of magnitude 1.80 T experiences a magnetic force of magnitude 7.60 10-1

3 N. What is the angle between the proton's velocity and the field
Physics
1 answer:
Hoochie [10]3 years ago
4 0

Hi there!

We can use the following equation for a point charge in a magnetic field:


\large\boxed{F_B = qv \times B}

F_B = Force due to magnetic field (7.6 × 10⁻¹³N)
q = Charge of particle (1.6 × 10⁻¹⁹ C)

v = velocity of particle (6.6 × 10⁶ m/s)

B = Magnetic field strength (1.8 T)

Or, without the cross product:
F_B = qvBsin\theta

θ = angle between particle's velocity and field

We can rearrange to solve for theta:
\frac{F_B}{qvB} = sin\theta\\\\\theta = sin^{-1} (\frac{F_B}{qvB})

Solve for theta:
\theta = sin^{-1} (\frac{7.6*10^{-13}}{(1.6*10^{-19})(6.6*10^6)(1.80)}) = \boxed{23.57^o}

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Answer:

hmax = 1/2 · v²/g

Explanation:

Hi there!

Due to the conservation of energy and since there is no dissipative force (like friction) all the kinetic energy (KE) of the ball has to be converted into gravitational potential energy (PE) when the ball comes to stop.

KE = PE

Where KE is the initial kinetic energy and PE is the final potential energy.

The kinetic energy of the ball is calculated as follows:

KE = 1/2 · m · v²

Where:

m = mass of the ball

v = velocity.

The potential energy is calculated as follows:

PE = m · g · h

Where:

m = mass of the ball.

g = acceleration due to gravity (known value: 9.81 m/s²).

h = height.

At  the maximum height, the potential energy is equal to the initial kinetic energy because the energy is conserved, i.e, all the kinetic energy was converted into potential energy (there was no energy dissipation as heat because there was no friction). Then:

PE = KE

m · g · hmax = 1/2 · m · v²

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hmax = 1/2 · v² / g

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