Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

Question: Two long straight thin wires with current lie against an equally long plastic cylinder, at radius $R = 20.0 cm$ from the cylinder’s central axis.
Figure 29-58a shows, in cross section, the cylinder and wire 1 but not wire 2. With wire 2 fixed in place, wire 1 is moved around the cylinder, from angle localid="1663154367897" $θ_{1} = 0 °$ to angle localid="1663154390159" $θ_{1} = 180 °$ , through the first and second quadrants of the xy coordinate system. The net magnetic field $\vec{B}$ at the center of the cylinder is measured as a function of $θ_{1}$ . Figure 29-58b gives the x component $B_{x}$ of that field as a function of $θ_{1}$ (the vertical scale is set by $B_{x s} = 6.0 μ T$ ), and Fig. 29-58c gives the y component (the vertical scale is set by $B_{y s} = 4.0 μ T$ ). (a) At what angle $θ_{2}$ is wire 2 located? What are the (b) size and (c) direction (into or out of the page) of the current in wire 1 and the (d) size and (e) direction of the current in wire 2?

a) The angle $θ_{2}$ at which wire 2 is located is either $+ 90 ° or - 90 °$ .

b) The size of the current in wire 1 = 4.0 A.

c) The direction of the current in wire 1= out of the page

d) The size of the current in wire 2 = 2.0 A

e) The direction of the current in wire 2 is into the page.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: The given values and the concept.

The distance between the center of the cylinder and the wires is $R = 20.0 c m$ .

The wire 2 is fixed in place.

The magnetic field at a point due to a current carrying wire is determined using the Biot-Savart law. The direction of the magnetic field is decided by the right-hand rule. The net magnetic field at the center of the cylinder is the vector sum of the magnetic fields due to wire 1 and wire 2. Using this, we can answer the above questions.

Formula:

$B = \frac{μ_{0} i}{2 π R}$

Step 2: (a) Calculate the angle θ2 at which wire 2 is located.

The information about the net magnetic field at the center of the cylinder is given in figures (b) and (c). Using this, we can write the equations as

$\vec{B} = \vec{B_{1}} + \vec{B_{2}}$

In the component form,

$B_{x} = B_{1 x} + B_{2 x} B_{y} = B_{1 y} + B_{2 y}$

$B_{x, 0} = B_{1 x, 0} + B_{2 x} B_{y, 0} = B_{1 y, 0} + B_{2 y} B_{x, 90} = B_{1 x, 90} + B_{2 x} B_{y, 90} = B_{1 y, 90} + B_{2 y} B_{x, 180} = B_{1 x, 180} + B_{2 x} B_{y, 180} = B_{1 y, 180} + B_{2 y}$

Using the values form the graph, we write

data-custom-editor="chemistry" $2 = 0 + B_{2 x} - 4 = B_{1 y, 0} + B_{2 y} 6 = B_{1 x, 90} + B_{2 x} 0 = 0 + B_{2 y} 2 = 0 + B_{2 x} 4 = B_{1 y, 180} + B_{2 y}$

From these equations, we can find

$B_{2 x} = 2 μ T$ and $B_{2 y} = 0$

Hence the wire 2 must be along the y axis direction, i.e., its position must be at $90 °$ or (i.e., $270 °$ )

To avoid the collision of two wires, wire 2 must be at either $+ 90 ° or - 90 °$ .

Hence, the angle $θ_{2}$ at which wire 2 is located is either $+ 90 ° or - 90 °$ .

Step 3: (b) Calculate the size of the current in wire 1

We have

$B_{1 y} = \frac{μ_{0} i_{1}}{2 π R} 4 \times 10^{- 6} = \frac{μ_{0} i_{1}}{2 π R} B_{1 y, 180} = 4 μ T$

$\begin{array}{rcl} i_{1} & = & \frac{4 \times 10^{- 6} \times 2 π \times 20 \times 10^{- 2}}{4 π \times 10^{- 7}} \\ i_{1} & = & 40 \times 10^{- 1} \\ = & 4.0 A \end{array}$

Hence, the size of the current in wire 1 = 4.0 A.

Step 4: (c) Calculate the direction of the current in wire 1

The y component of the magnetic field is positive when wire 1 is at $180 °$ . This is possible only if the current in the wire 1 is out of the page. All the other equations are also valid for this direction of the current.

Hence, the direction of the current in wire 1= out of the page

Step 5: (d) Calculate the size of the current in wire 2

We have

$B_{2 x} = 2 μ T B_{2 x} = \frac{μ_{0} i_{2}}{2 π R} 2 \times 10^{- 6} = \frac{μ_{0} i_{2}}{2 π R}$

$\begin{array}{rcl} i_{2} & = & \frac{2 \times 10^{- 6} \times 2 π \times 20 \times 10^{- 2}}{4 π \times 10^{- 7}} \\ i_{2} & = & 20 \times 10^{- 1} \\ = & 2.0 A \end{array}$

Hence, the size of the current in wire 2 = 2.0 A.

Step 6: (e) Calculate the direction of the current in wire 2

The x component of the current in wire 2 is positive. Hence as seen in part (a), if the wire is at $- 90 °$ , the current must be going into the page

Hence, the direction of the current in wire 2 is into the page.

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: The given values and the concept.

Step 2: (a) Calculate the angle θ2 at which wire 2 is located.

Step 3: (b) Calculate the size of the current in wire 1

Step 4: (c) Calculate the direction of the current in wire 1

Step 5: (d) Calculate the size of the current in wire 2

Step 6: (e) Calculate the direction of the current in wire 2

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: The given values and the concept.

Step 2: (a) Calculate the angle θ2 at which wire 2 is located.

Step 3: (b) Calculate the size of the current in wire 1

Step 4: (c) Calculate the direction of the current in wire 1

Step 5: (d) Calculate the size of the current in wire 2

Step 6: (e) Calculate the direction of the current in wire 2

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