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3 years ago

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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 magne

tic field at the following (xx, yy, zz) positions?
a. (1 cm , 0 cm, 0 cm)
b. (0 cm, 0 cm, 2 cm)

1 answer:

Artist 52 [7]3 years ago

6 0

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

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

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

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