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zepelin [54]
3 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]3 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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Answer: a) 3.85 days

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Explanation:-

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = ?

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First we have to calculate the rate constant, we use the formula :

Now put all the given values in above equation, we get

k=\frac{2.303}{3}\log\frac{100}{58}

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t_{\frac{1}{2}}=\frac{0.693}{0.18}=3.85days

Thus half-life of radon-222 is 3.85 days.

b) Time taken for the sample to decay to 15% of its original amount:

where,

k = rate constant  = 0.18days^{-1}

t = time taken for decomposition  = ?

a = let initial amount of the reactant  = 100 g

a - x = amount left after decay process  = \frac{15}{100}\times 100=15g

t=\frac{2.303}{0.18}\log\frac{100}{15}

t=10.54days

Thus it will take 10.54 days for the sample to decay to 15% of its original amount.

3 0
3 years ago
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<span>reflection, rotation, translation</span>
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deff fn [24]

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8 0
3 years ago
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When the distance between two interacting objects doubles, the gravitational force is
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The gravitational force will be one quarter.

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5 0
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