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Question No.34
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Linear charge density of wire is 8.85 uC/m Radius and height of the cylinder are 3m and 4 m. Then find the flux passing through the cylinder
wire
th
+
+
4 m
3 m
O 5 x 100 volt-m
O 3 x 106 volt-m
4 x 106 volt-m
O None
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Answer 1

Answer:

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Answer 2

Answer:

The flux passing through the wire is

$5 \times 10 {}^{6} \: volt - m$

Explanation:

Given that,

A wire of certain linear charge density passes through a cylinder.

Also given that,

The radius of the cylinder is 3m and the height of the cylinder is 4m.

$= > r = 3m \\ = > h = 4m$

To find the electric flux we need to find the length of wire inside the cylinder.We find the length of wire inside cylinder in the following way

$l = \sqrt{ {r}^{2} + {h}^{2} } \\ = \sqrt{ {3}^{2} + {4}^{2} } = 5m$

Hence the length of wire inside the cylinder is 5m.

Given that,the charge density of the wire is

$\lambda = 8.85 \times 10 {}^{ - 6}C/m$

The charge inside the wire is product of charge density of wire and its length

$Q_{in}=\lambda \: l$

Substituting the known values in the above relation we get

$Q_{in}=8.85 \times 10 {}^{ - 6} \times 5 \: C$

The flux passing through the wire is the ratio of charge passing through the wire and electric coefficient

$flux = \frac{Q_{in}}{\epsilon_{0}}$

The value of the coefficient is

$\epsilon_{0}=8.85\times10^{-12} C^{2}/Nm^{2}$

Substituting this value in the relation of flux we get

$flux = \frac{8.85 \times 10 {}^{ - 6} \times 5}{8.85 \times 10 {}^{ - 12} } \\ = 5 \times 10 {}^{6} \: volt - m$

Therefore,the flux of wire passing through the cylinder is

$5 \times 10 {}^{6} \: volt - m$

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