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**Hint**Derive the relation between the area and the diameter and substitute this in the resistance formula to find the resistance per unit length of the solenoid. From the magnetic flux density formula, find the current and the diameter relation. Substitute both in the formula of the heat dissipated to find the relation between the heat dissipated and the diameter of the solenoid.

Useful formula:

The formula of the heat dissipated per unit length is given by

$H = R{I^2}$

Where $H$ is the heat dissipated, $R$ is the resistance per unit length of the solenoid and $I$ is the current through the solenoid.

**Complete step by step answer**

It is given that there are the two cylinders of the fixed radii. The space between these two are wound by the solenoid that provides the magnetic flux density. It is known that the cross sectional area that is to be filled by the solenoid depends on the square of the dimeter, \[{d^2}\] . Hence $n\alpha {d^{ - 2}}$ . The resistance varies inversely to that of the cross sectional area , hence $R\alpha {d^{ - 2}}$ . By using this, we can find the resistance per unit length of the solenoid. It is calculated as $R\alpha n{d^{ - 2}}$

We know that the $n\alpha {d^{ - 2}}$ , and hence

$R\alpha {d^{ - 4}}$

We know that the magnetic flux density is directly proportional to the product of the current and the number of turns.

$B\alpha nI$

From the above step, we can write

$I\alpha \dfrac{1}{n}\alpha {n^{ - 1}}$

Substituting the value of the number of turns in the above equation, we get

$I\alpha {d^2}$

Let us use the formula of the heat dissipated,

$H = R{I^2}$

Substituting the known value in it, we get

$H = {d^{ - 4}}{\left( {{d^2}} \right)^2} = 1$

The above answer shows that the heat dissipated is independent of the diameter of the solenoid.

**Hence the option (C) is correct.**

**Note**The heat dissipation in the solenoid mainly depends on the current flowing through the solenoid, resistance of the wire. If the current through the wire is opposed by the more resistance of the wire, then the heat dissipation by that solenoid is also more.

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