A long straight conducting rod carries a current I with a non-uniform current density J = ar2, and has a radius R. The value of the constant

Question

A long straight conducting rod carries a current I with a non-uniform current density J = ar2, and has a radius R. The value of the constant is 28.5 A/mm4 and the radius of the rod is 5.20 mm. Determine the magnitude of the magnetic field at the following points.(a) r1 = R/2(b) r2 = 2R

in progress 0
Kim Cúc 4 years 2021-07-25T18:15:47+00:00 1 Answers 32 views 0

Answers ( )

  1. huyenthanh
    0
    2021-07-25T18:17:12+00:00
    Reply

    Answer:

    Following are the solution to this question:

    Explanation:

    Given value:

    Current density J = ar^2

    value of the constant is=  28.5 \ \frac{A}{mm^4}

    radius = 5.20 mm

    magnetic permeability \mu = 4\pi \times 10^7 \ \frac{N}{A^2}

    calculating the area element for the straight circular conduction rod
    :

    d_A=2\pi r dr

    Calculating the current, which is carrying in rod

    dI = \int dA \vec{J}

    Calculating the above equation with the limit value that is 0 to r.

    I^{1}=\int_{0}^{r} ar^2 \times 2 \pi \cdot r d_r

        =2\pi a\int_{0}^{r} r^3 d_r \\\\=\frac{\pi ar^4}{2}

    The calculated current value which is carried by the rod is \boxed{\frac{\pi ar^4}{2}}

    In option (a)

    calculating the magnitude of the magnetic field at the point r_1= \frac{R}{2}

    \to B=\frac{\mu_{0} I^{1}}{2 \pi r}\\\\\to B \times 2\pi r= \mu_{0} (\frac{\pi a}{2}r^4)\\\\\to B \times 2\pi = \mu_{0} (\frac{\pi a}{2}r^3)\\\\\to B= \frac{\mu_{0}}{4}r^3 a\\\\

    Substituting the above value:\frac{R}{2} \ for \ r

    B= \frac{\mu_{0}}{4}(\frac{R}{2})^3 a

    B= \frac{4 \pi \times 10^{-7}}{4}(\frac{5.2 \times 10^{-3}}{2})^3 \frac{28.5}{10^{-12}}

       = \frac{4 \pi \times 10^{-7}}{4} \times \frac{140.608 \times 10^{-9}}{8} \times \frac{28.5}{10^{-12}}\\\\= \frac{ 3.14 \times 10^{-7}}{1} \times \frac{140.608 \times 10^{-9}}{8} \times \frac{28.5}{10^{-12}}\\\\=\frac{1572.87624 \times 10^{-16}}{ 10^{-12}}\\\\=0.157 \ \ T

    Thus, the magnitude of the magnetic field at the point r_1 =\frac{R}{2} is \boxed{0.157 \ T}

    In option (b)

    In this, we calculate the magnitude of the magnetic field at the point r_2= 2R

    \to B=\frac{\mu_{0} I^{1}}{2 \pi r} \\\\\to B \times 2\pi r= \mu_{0} (\frac{\pi a}{2}R^4)\\\\\to B \times 2\pi 2R = \mu_{0} (\frac{\pi a}{2}R^3)\\\\\to B= \frac{\mu_{0}}{8}R^3 a\\\\

    Substituting the values  

    B= \frac{4 \pi \times 10^{-7}}{8}(5.2 \times 10^{-3})^3(\frac{28.5}{10^{-12}})

       = \frac{4 \times 3.14 \times 10^{-7}}{8} \times (5.2)^3 \times (10^{-3})^3 \times\frac{28.5}{10^{-12}}\\\\= \frac{ 3.14 \times 10^{-7}}{2} \times 140.608 \times 10^{-9} \times\frac{28.5}{10^{-12}}\\\\= 6291.50495 \times 10^{-4}\\\\= 0.629 \ \ or \ \ 0.63\\

    Thus, the magnitude of the magnetic field at the point r_2 = 2R is \boxed{0.63 \ \ T}

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )