Question

An electron moves along the z-axis with vz=4.1Ã107m/s. As it passes the origin, what are the strength and direction of the magnetic field at the following (xx, yy, zz) positions?
a. (1 cm , 0 cm, 0 cm)
b. (0 cm, 0 cm, 2 cm)

Answer 1

Answer:

a

The value of magnetic field is $B_a=6.56*10^{-19} T$

In the direction of the positive x - axis

b

The value of magnetic field is $B_a=6.56*10^{-19} T$

In the direction of the positive z - axis

Explanation:

From the question we are told that

         The velocity of the electron is $v_z = 4.1*10^{7}m/s$

Considering the first position

    The equation for the magnetic field is

                    $\= B = \frac{\mu_0}{ 4 \pi} \frac{q \=v * \= r}{r^2}$

 Now $r = \sqrt{i^2 + j^2 + k^2}$

  substituting values

           $r = \sqrt{1^2 +0^2 +0^2}$

              $= 1$

          $\= r = \frac{\r r}{r}$

        $\r r = 1 i + 0j + 0k$

Therefore $\= r = \frac{1i + 0j + 0k}{1}$

                   $= i$

So   Substituting  $4 \pi *10^ {-7}$ for $\mu_o$ , $1.602 *10^{-19}$ for q

            $\= B_a = \frac{4\pi *10^{-7}}{4 \pi} \frac{1.602 *10^{-19} * 4.1*10^{7} * i}{1^2}$

                 $\= B_a=6.56*10^{-19} T (i)$

Considering the second  position

    Here

               $r = \sqrt{0^2 + 0^2 + 2^2}$

                  $=2$

              $\= r = \frac{\r r}{r}$

              $\r r = 0 i + 0j + 2k$

            $\= r = \frac{0i + 0j + 2k}{2}$

               $= k$

 So   Substituting  $4 \pi *10^ {-7}$ for $\mu_o$ , $1.602 *10^{-19}$ for q

            $\= B_a = \frac{4\pi *10^{-7}}{4 \pi} \frac{1.602 *10^{-19} * 4.1*10^{7} * k}{1^2}$

                 $\= B_a=6.56*10^{-19} T (k)$