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larisa [96]
4 years ago
15

A small current element carrying a current of I = 1.00 A is placed at the origin given by d → l = 4.00 m m ^ j Find the magnetic

field, d → B , at the locations specified. Enter the correct magnitude and select the direction from the list. If the direction is negative, indicate this by entering the magnitude as a negative number. What is the magnitude and direction of d → B on the x ‑axis at x = 2.50 m ? magnitude: T Direction: ^ k ^ j ^ i What is the magnitude and direction of d → B on the z ‑axis at z = 5.00 m ?
Physics
1 answer:
xxTIMURxx [149]4 years ago
7 0

Answer:

the magnitude and direction of d → B on the x ‑axis at x = 2.50 m is -6.4 × 10⁻¹¹T(Along z direction)

the magnitude and direction of d → B on the z ‑axis at z = 5.00 m is 1.6 × 10⁻¹¹T(Along x direction)

Explanation:

Use Biot, Savart, the magnetic field

d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}

Given that,

i = 1.00A

d → l = 4.00 m m ^ j

r = 2.5m

Displacement vector is

\bar{r}=x\hat i+y\hat j+z \hat k\\

\bar{r}= (2.5m) \hat i +(0m)^2 + (0m)^2

 =2.5m

on the axis of x at x = 2.5

r = \sqrt{(2.5)^2 + (0)^2 + (0)^2}

r = 2.5m

And unit vector

\hat r =\frac{\bar{r}}{r}

= \frac{2.5 \hat i}{2.5}\\\\= 1\hat i

Therefore, the magnetic field is as follow

d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}

d\bar{B} = \frac{(10^-^7)(1)(4\times10^-^3j\times i}{(2.50)^2} \\\\d\bar{B} = -6.4\times10^{-11} T

(Along z direction)

B)r = 5.00m

Displacement vector is

\bar{r}=x\hat i+y\hat j+z \hat k\\

\bar{r}= (5.00m) \hat i +(0m)^2 + (0m)^2

 =5.00m

on the axis of x at x = 5.0

r = \sqrt{(5.00)^2 + (0)^2 + (0)^2}

r = 5.00m

And unit vector

\hat r =\frac{\bar{r}}{r}

= \frac{5.00 \hat i}{5.00}\\\\= 1\hat i\\

Therefore, the magnetic field is as follow

d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}

d\bar{B} = \frac{(10^-^7)(1)(4\times10^-^3j\times i}{(5.00)^2} \\\\d\bar{B} = 1.6\times10^{-11} T

(Along x direction)

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