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2 years ago

If the resulting trajectory of the charged particle is a circle, what is ω, the angular frequency of the circular motion?

Physics

1 answer:

Komok [63]2 years ago

7 0

For this answer you need to have in mind the following formulas:

Fmag = qvB

Also have in mind that Centripetal acceleration is: ar = v2/R

centripetal force will be :Fc = mv2/R

And velocity is: v = ωR

Remember that the magnetic force is the centripetal force then:

Fmag = Fc
qvB = mv2/R
qB = mv/R

Next you need to substitute for velocity:

qB = mωR/R
qB = mω

ω = qB/m

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A car was at a 30km marker and began to drive. The car passed a 90km marker and the time had changed 30

dolphi86 [110]

Answer:

i think it's 2km pm

Explanation:

2km x 30 60.. start was 30, and now your at 90.. we had to determine how much time it took.. so 2 is the average.. or atleast per minute and sorry it i still didnt answer ur question lol im just trynna help

7 0

2 years ago

What happens to the magnitude of the gravitational force as the distance between two bodies increase?

Anna [14]

Answer:

The magnitude of the force will decrease

Explanation:

The gravitational force is one of the four fundamental forces of nature. It is an attractive force exerted between every object having mass.

Its magnitude is given by the equation:

$F=\frac{Gm_1 m_2}{r^2}$

where

G is the gravitational constant

m1 is the mass of the first object

m2 is the mass of the second object

r is the separation between the objects

As we see from the equation, the magnitude of the gravitational force is inversely proportional to the square of the distance between the objects:

$F\propto \frac{1}{r^2}$

Therefore, this means that as the distance between two bodies increases, the gravitational force will decrease.

7 0

3 years ago

A proton is projected toward a fixed nucleus of charge Ze with velocity vo. Initially the two particles are very far apart. When

11111nata11111 [884]

Answer:

The value is $R_f = \frac{4}{5} R$

Explanation:

From the question we are told that

The initial velocity of the proton is $v_o$

At a distance R from the nucleus the velocity is $v_1 = \frac{1}{2} v_o$

The velocity considered is $v_2 = \frac{1}{4} v_o$

Generally considering from initial position to a position of distance R from the nucleus

Generally from the law of energy conservation we have that

$\Delta K = \Delta P$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K = K__{R}} - K_i$

=> $\Delta K = \frac{1}{2} * m * v_1^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * (\frac{1}{2} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta P = P_f - P_i$

Here $P_i$ is zero because the electric potential energy at the initial stage is zero so

$\Delta P = k * \frac{q_1 * q_2 }{R} - 0$

$\frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R} - 0$

=> $\frac{1}{2} * m *v_0^2 [ \frac{1}{4} -1 ] = k * \frac{q_1 * q_2 }{R}$

=> $- \frac{3}{8} * m *v_0^2 = k * \frac{q_1 * q_2 }{R} ---(1 )$

Generally considering from initial position to a position of distance $R_f$ from the nucleus

Here $R_f$ represented the distance of the proton from the nucleus where the velocity is $\frac{1}{4} v_o$

Generally from the law of energy conservation we have that

$\Delta K_f = \Delta P_f$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K_f = K_f - K_i$

=> $\Delta K_f = \frac{1}{2} * m * v_2^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * (\frac{1}{4} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * \frac{1}{16} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance $R_f$ from the nucleus , this is mathematically represented as

$\Delta P_f = P_f - P_i$