
Find the resistance of the coil of length l, number of turns per unit length n, the diameter of the wire ‘d’, radius of the coil R, and resistivity p?

Answer
484.2k+ views
Hint: The question is based on the concept of the resistivity of the material. So, we will be making use of the resistivity of the material formula to find the resistance of the coil. We will substitute the parameters of the length and area of the material in terms of the length and area of the coil.
Formula used:
\[R=\dfrac{\rho l}{A}\]
Complete step-by-step solution
The resistivity of the material is the ability to resist the flow of current through the material.
The formula for computing the resistivity of the material is given as follows.
\[R=\dfrac{\rho l}{A}\]
Where R represents the resistance of the material, \[\rho \] represents the resistivity of the material, l represents the length of the material and A represents the area of the material.
From the given, we have the data as follows.
The length of the coil is l.
The number of turns per unit length is n.
The diameter of the wire is d.
The radius of the coil is R.
The resistivity of the coil is \[\rho \].
The total number of turns in the coil is, \[N=nl\]
The length of the wire is, \[L=N(2\pi R)\]
Where length is the total number of turns times the circumference of the coil.
Therefore, the length of the wire is,
\[\begin{align}
& L=nl(2\pi R) \\
& L=2\pi nlR \\
\end{align}\]
The area of the coil is equal to the area of the circle, so, we have,
The area of the coil is, \[A=\pi {{r}^{2}}\]
\[A=\dfrac{\pi {{d}^{2}}}{4}\]
Substitute the expressions of these parameters in the equation of the resistance of the material. So, we get,
The resistance of the coil to be equal to,
\[\begin{align}
& R=\dfrac{\rho l}{A} \\
& R=\dfrac{\rho \times (2\pi nlR)}{\dfrac{\pi {{d}^{2}}}{4}} \\
& R=\dfrac{8nlR\rho }{{{d}^{2}}} \\
\end{align}\]
\[\therefore \] The expression for the resistance of the coil obtained is \[R=\dfrac{8nlR\rho }{{{d}^{2}}}\].
Note: This is a direct question. The formulae used in the above explanation should be known to solve these types of derivational questions. The length and area of the material in terms of the length and area of the coil substitution should be known.
Formula used:
\[R=\dfrac{\rho l}{A}\]
Complete step-by-step solution
The resistivity of the material is the ability to resist the flow of current through the material.
The formula for computing the resistivity of the material is given as follows.
\[R=\dfrac{\rho l}{A}\]
Where R represents the resistance of the material, \[\rho \] represents the resistivity of the material, l represents the length of the material and A represents the area of the material.
From the given, we have the data as follows.
The length of the coil is l.
The number of turns per unit length is n.
The diameter of the wire is d.
The radius of the coil is R.
The resistivity of the coil is \[\rho \].
The total number of turns in the coil is, \[N=nl\]
The length of the wire is, \[L=N(2\pi R)\]
Where length is the total number of turns times the circumference of the coil.
Therefore, the length of the wire is,
\[\begin{align}
& L=nl(2\pi R) \\
& L=2\pi nlR \\
\end{align}\]
The area of the coil is equal to the area of the circle, so, we have,
The area of the coil is, \[A=\pi {{r}^{2}}\]
\[A=\dfrac{\pi {{d}^{2}}}{4}\]
Substitute the expressions of these parameters in the equation of the resistance of the material. So, we get,
The resistance of the coil to be equal to,
\[\begin{align}
& R=\dfrac{\rho l}{A} \\
& R=\dfrac{\rho \times (2\pi nlR)}{\dfrac{\pi {{d}^{2}}}{4}} \\
& R=\dfrac{8nlR\rho }{{{d}^{2}}} \\
\end{align}\]
\[\therefore \] The expression for the resistance of the coil obtained is \[R=\dfrac{8nlR\rho }{{{d}^{2}}}\].
Note: This is a direct question. The formulae used in the above explanation should be known to solve these types of derivational questions. The length and area of the material in terms of the length and area of the coil substitution should be known.
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