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zepelin [54]
4 years ago
15

A hollow cylinder with an inner radius of 4 mm and an outer radius of 25 mm conducts a 4-A current flowing parallel to the axis

of the cylinder. If the current density is uniform throughout the wire, what is the magnitude of the magnetic field at a point 17 mm from its center? (μ 0 = 4π × 10-7 T · m/A)
Physics
1 answer:
ivann1987 [24]4 years ago
5 0

Answer:

2.1 × 10⁻⁵ T

Explanation:

Given:

Inner radius, r = 4 mm = 0.004 m

Outer radius, R = 25 mm = 0.025 m

Current, I = 4 A

Distance of the point from the center, a = 17 mm = 0.017 m

μ₀ = 4π × 10⁻⁷ T·m/A

Now,

For the hollow cylinder magnetic field (B) is given as:

B=\frac{\mu_0 I(a^2-r^2)\textup{}}{2\pi a(R^2-r^2)\textup{}}

on substituting the respective values, we get

B=\frac{4\pi\times10^{-7}\times4(0.017^2-0.004^2)\textup{}}{2\pi\times0.017(0.025^2-0.004^2)\textup{}}

or

B = 2.1 × 10⁻⁵ T

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If it takes 35J to lift up an object to some height, then how much gravitational potential energy will that object have?
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State the formula for calculating power
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3 years ago
In a science museum, a 110 kg brass pendulum bob swings at the end of a 13.9 m -long wire. the pendulum is started at exactly 8:
saw5 [17]

The number of oscillations completed by the pendulum is 2736.

The amplitude of the pendulum is 3.47 m.

The given motion is an underdamped motion. So its frequency will be similar to that of a simple harmonic motion.

The frequency of oscillation is defined as the number of oscillations completed in unit time. It is calculated using the formula.

f=(1/2π)*√(l/g)

where f is the frequency, l is the length of the pendulum, and g is the acceleration due to gravity.

Given the length of the wire l=13.9 m and acceleration due to gravity g=9.8 m/s^2. The frequency of oscillation is:

f=(1/(2*3.14)) * √(13.9/9.8)

f=0.19 Hz (approximately)

Since the pendulum started oscillating at 8:00 am, 4 hours has been passed when it shows 12:00 pm. So time t=4 hours or t=4*3600. Hence t=14400 s. The total number of oscillations is then given by the formula,

n=ft

where n is the number of oscillations.

n=0.19*14400=2736.

In damping motion, the amplitude of the pendulum decreases with time. The amplitude of the pendulum is given by the formula,

A' = A exp (-b*t)

where A' is the amplitude after time t, A is the initial amplitude, b is the damping constant, and t is the time.

Here A=1.2 m, b=0.010 kg/s and t=14400 s.

A' = 1.2 exp (-0.010*14400)

A'=3.47 m (approximately)

Learn more about amplitude.

brainly.com/question/21632362

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5 0
2 years ago
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