Chapter 4: Q11E (page 948)

A long, straight wire lies along the z-axis and carries a 4.00-A current in the +z-direction. Find the magnetic field (magnitude and direction) produced at the following points by a 0.500-mm segment of the wire cantered at the origin: (a) x = 2.00 m, y = 0, z = 0; (b) x = 0, y = 2.00 m, z = 0; (c) x = 2.00 m, y = 2.00 m, z = 0; (d) x = 0, y = 0, z = 2.00 m,

Short Answer

Expert verified

(a) the magnetic field (magnitude and direction) produced at the following points by a 0.500 - mm segment of the wire cantered at the origin $x = 2.00 m, y = 0, z = 0 is (50.0 pT) \hat{i}$

(b) the magnetic field (magnitude and direction) produced at the following points by a 0.500-mm segment of the wire cantered at the origin $x = 0, y = 0, 2.00 m, z = 0 is (50.0 pT) \hat{i}$

(c) the magnetic field (magnitude and direction) produced at the following points by a 0.500 - mm segment of the wire cantered at the origin $x = 2.00 m, y = 0, 2.00 m, z = 0 is (17.7 pT) (- \hat{i} + \hat{j})$

(d) the magnetic field (magnitude and direction) produced at the following points by a segment of the wire cantered at the origin $x = 0, y = 0, z = 2.00 m is 0$

Step by step solution

Direction of at point in the - x direction

Given the field point having co-ordinates (2.00,0,0)m $s o \hat{r} = (2.00 m) \hat{i}$ and

$\hat{B} = \frac{μ_{0}}{4 π} \frac{l d \hat{l} \times \hat{r}}{r^{2}} = \frac{μ_{0}}{4 π} l d l r^{2} \hat{j} = (10^{- 7} T . m / A) \frac{(4.00 A) (0.500 \times 10^{- 3} m)}{2.00 m)^{2} \hat{j}} = (50.0 p T) \hat{j}$

The direction of at the point in the direction is

For the points (0,2.00,0)m

For the field point has the coordinates $(0, 2.00, 0) m s o \hat{r} = (2.00 m) \hat{j}$

Now, $\hat{B} = \frac{μ_{0}}{4 π} \frac{l d \hat{l}}{r^{2}} = - \frac{μ_{0}}{4 π} \frac{l D \hat{l}}{r^{2}} \hat{l}$

$= - (10^{- 7} T . m / A) \frac{4.00 A (0.500 \times 10^{- 3} m)}{{(2.00)}^{2}} \hat{i} = - (50.0 p T) \hat{i}$

The direction of $\hat{B}$ at this point is in the - x direction

For xy-planes

For the field point the radial position

$\hat{r} = (2 m) (\hat{i} + j) {andr}^{2} = 8.00 m^{2} \hat{B} = \frac{μ_{0}}{4 π} \frac{ld \hat{l} \times \hat{r}}{r^{2}} = \frac{(10^{- 7} T . m / A) (4.00 A) (0.500 \times 10^{- 3} m)}{8.00 m^{2}} (\hat{- i} + j) = (17.7 pT) (- \hat{i} + \hat{j}))$

The vector $\hat{B}$ lies in the xy planes and makes an angle of $45.0 °$ with the -x direction.

The field point has the coordinates (0,02.00) m , hence the cross products

$d \hat{l} \times \hat{r} = dl (\hat{k}) \times \hat{k} = 0 and thus \hat{B} = 0$

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Short Answer

Step by step solution

Direction of at point in the - x direction

For the points (0,2.00,0)m

For xy-planes

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