Q3P - Fundamentals Of Physics

Question

A Gaussian surface in the shape of a right circular cylinder with end caps has a radius of 12.0 cm and a length of 80.0 cm. Through one end there is an inward magnetic flux $\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{\mu }\mathit{W}\mathit{b}$. At the other end, there is a uniform magnetic field 1.60 mT, normal to the surface and directed outward. What are the (a) magnitude and (b) direction (inward or outward) of the net magnetic flux through the curved surface?

Step-by-Step Solution

Answer

1. The magnitude of the magnetic flux is $47.4 \mu Wb$.
2. The direction of the magnetic flux is inward.

Step 1: Given

• The radius of the caps of the cylinder r = 12.0 cm = 0.12 m
• Length of the cylinder L = 80 cm = 0.8 m
• At the first cap 1, inward magnetic flux, ${\phi }_1=-24 \mu Wb=-25\times {10}^{-6}Wb$
• At the other cap 2, the magnetic field $B_2=1.60 mT=1.60\times {10}^{-3} T$

Step 2: Determining the concept

Applying Gauss law for magnetism, write for magnetic flux through the circular cylinder. Inserting given values in it, find the magnitude and direction of the magnetic flux through the curved part.

The formula is as follows:

Gauss law for any closed Gaussian surface, ${\phi }_b=\oint B.dA=0$

Step 3: (a) Determining the magnitude of the magnetic flux through the curved part

In the given closed surface, that is, the circular cylinder, G ${\phi }_{cap 2}=0.0723456\times {10}^{-3} Wb$. Gauss law gives that net magnetic flux through any closed surface is zero.

$$\begin{gathered} {\phi }_b=\oint B.dA=0 \\ {\phi }_b={\phi }_{cap 1}+{\phi }_{cap 2}+{\phi }_{curved surface}=0..........\left(1\right) \end{gathered}$$

It is known, ${\phi }_{cap 1}=-24\times {10}^{-6} Wb$

$$\begin{gathered} {\phi }_{cap 2}=\oint B.dA=B_{cap 2}{\mathrm{\pi r}}^2 \\ {\mathrm{\phi }}_{\mathrm{cap} 2}=1.60\times {10}^{-3}\times 3.14\times {\left(0.12\right)}^2 \\ {\mathrm{\phi }}_{\mathrm{cap} 2}=72.4\times {10}^{-6}\mathrm{Wb} \end{gathered}$$

Using this in equation 1,

$\begin{array}{rcl}-25\times {10}^{-6}+72.4\times {10}^{-6}+{\phi }_{curved surface}& =& 0\\ 47.4\times {10}^{-6}+{\phi }_{curved surface}& =& 0\\ {\phi }_{curved surface}& =& -47.4\times {10}^{-6}Wb\\ {\phi }_{curved surface}& =& -47.4 \mu Wb\\ \left|{\phi }_{curved surface}\right|& =& 47.4 \mu Wb\end{array}$

Therefore, the magnitude of magnetic flux through the curved surface is $\begin{array}{rcl}& & 47.4 \mu Wb\end{array}$.

Step 4: (b) Determining the direction of the magnetic flux

The direction of magnetic flux through a curved surface is inwards since its negative.

Applying Gauss law for magnetism, find the magnetic flux through one part using the magnetic flux through the other parts of the closed surface.