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Amiraneli [1.4K]

3 years ago

5

A uniform electric field of magnitude 110 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.

64 T perpendicular to the electric field also exists in this region. A beam of positively charged particles travels into the region. Determine the speed of the particles at which they will not be deflected by the crossed electric and magnetic fields. (Assume the beam of particles travels perpendicularly to both fields.)

2 answers:

Brut [27]3 years ago

4 0

Answer: 1.71*10^5 m/s

Explanation:

Given

Magnitude of electric field, E = 110 kV/m

Magnetic field, B = 0.64 T

The forces acting on the charged particle.

Due to electric field = qE, and it's an upward one

Due to magnetic field = qVB, this is a downwards one judging from right hand screw rule

Since the two forces are equal and opposite, then

qE = qVB

E = VB and then,

V = E / B

V = 110*10^3 / 0.64

V = 171.875*10^3 m/s

Therefore, we can say, the speed of the particle at which they will not be deflected is 1.72*10^5 m/s

Marizza181 [45]3 years ago

3 0

Answer:

1.7×10^5 ms-1

Explanation:

From

qE= qvB

q= charge on the electron

E = electric field

v= velocity

B= magnetic field

E= vB

v= E/B= 110×10^3/0.6

v= 1.7×10^5 ms-1

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a

The New angular speed is $w_f = 2.034 rad/s$

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From the we are told that masses are 1.08 m from the axis of rotation, this means that

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The angular speed $w = 0.770 \ rad/sec$

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$w_i$ is the initial angular speed

$I_i$ is the initial moment of inertia

$I_f$ is the final moment of inertia

Moment of inertia is mathematically represented as

$I = m r^2$

Where I is the moment of inertia

m is the mass

r is the radius

So the Initial moment of inertia is given as

$I_i = moment \ of \ inertia \ of\ the \ two \ mass \ + 3.25 \ kg \cdot m^2$

$I_i = 2m r^2 + 3.25$

The multiplication by is because we are considering two masses

$I_i = 2 [(3.09)(1.08)^2] +3.25 = 10.46 kg \cdot m^2$

So the final moment of inertia is given as

$I_f = 2[(3.09)(0.34)^2] +3.25 = 3.96 \ kg \cdot m^2$

Substituting these values into equation 1

$w_f = [\frac{10.46}{3.96} ] * 0.77 = 2.034 \ rad/sec$

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Now considering the kinetic energy before the masses are pulled in,

$KE_i = \frac{1}{2} * I_i * w^2_i$

The Moment of inertia would be $I_i = 10.46 \ Kg \cdot m^2$

The Angular speed would be $w_i = 0.77 \ rad/s$

Now substituting these value into the equation above

$KE_i = \frac{1}{2} * (10.46) * (0.770)^2 = 3.101 J$

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$KE_f = \frac{1}{2} * I_f * w^2_f$

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The Angular speed would be $w_f = 2.034 \ rad/s$

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