Question

A proton moves along the x-axis with vx=1.0×107 m/s. As it passes the origin, what are the strength and direction of the magnetic field at the (x, y, z) positions (a) (1 cm, 0 cm, 0 cm), (b) (0 cm, 1 cm, 0 cm), and (c) (0 cm, −2 cm, 0 cm)?

Answer 1

Answer:

The magnetic field in x axis is 0.

The magnetic field in y axis is $1.6\times10^{-15}\ T$.

The magnetic field in y axis is $0.4\times10^{-15}\ T$.

Step-by-step explanation:

Given that,

Speed $v_{x} =1.0\times10^{7}\ m/s$

(a). For position (1,0,0),

A proton moves along the x-axis and the position of the particle is in x axis so the angle is 0°

We need to calculate the magnetic field

Using formula of magnetic field

$B=\dfrac{\mu_{0}qv\sin\theta}{4\pi r^2}$

$B=\dfrac{4\pi\times10^{-7}\times1.6\times10^{-19}\times1.0\times10^{7}\sin0^{\circ}}{4\pi\times(1\times10^{-2})^2}$

$B=0\ T$

(b). For position (0,1,0),

A proton moves along the x-axis and the position of the particle is in y axis so the angle is 90°

Using formula of magnetic field

$B=\dfrac{\mu_{0}qv\sin\theta}{4\pi r^2}$

$B=\dfrac{4\pi\times10^{-7}\times1.6\times10^{-19}\times1.0\times10^{7}\sin90^{\circ}}{4\pi\times(1\times10^{-2})^2}$

$B=1.6\times10^{-15}\ T$

(c).  For position (0,-2,0),

A proton moves along the x-axis and the position of the particle is in -y axis so the angle is 270°

Using formula of magnetic field

$B=\dfrac{\mu_{0}qv\sin\theta}{4\pi r^2}$

$B=\dfrac{4\pi\times10^{-7}\times1.6\times10^{-19}\times1.0\times10^{7}\sin270^{\circ}}{4\pi\times((-2)\times10^{-2})^2}$

$B=-0.4\times10^{-15}\ T$

Hence, The magnetic field in x axis is 0.

The magnetic field in y axis is $1.6\times10^{-15}\ T$.

The magnetic field in y axis is $0.4\times10^{-15}\ T$.

Answer 2

We want to find the magnetic field generated by a moving proton at different points.

a) B = 0

b) B =  $B = \frac{\mu_0}{4*\pi} \frac{e* (0 cm, 0cm, 1.0\times 10^5 m^2/s) }{1cm}$

c) B =  $B = \frac{\mu_0}{4*\pi} \frac{e* (0 cm, 0cm, -2.0\times 10^5 m^2/s) }{1cm}$

The general formula is:

$B = \frac{\mu_0}{4*\pi} \frac{q*(vxr)}{|r|}$

Where:

v×r is the cross product between the velocity and each position.

q = e, the charge of an electron, but positive.

v = (1.0×10^7 m/s, 0 , 0)

Now we just use the formula for each case:

a) r =  (1 cm, 0 cm, 0 cm)

Notice that this is parallel to the velocity, thus:

v×r = 0

Then B = 0.

b) r =  (0 cm, 1 cm, 0 cm)

|r| = 2cm

v×r = (0 cm, 0cm, 1.0×10^7 m/s*1cm) =  (0 cm, 0cm, 1.0×10^5 m^2/s)

Then the magnetic field is just:

$B = \frac{\mu_0}{4*\pi} \frac{e* (0 cm, 0cm, 1.0\times 10^5 m^2/s) }{1cm}$

c) This time we have:

r = (0 cm, −2 cm, 0 cm)

then:

|r| = 2cm

And:

v×r = (0 cm, 0cm, 1.0×10^7 m/s*-2cm) =  (0 cm, 0cm, -2.0×10^5 m^2/s)

Then the magnetic field is just:

$B = \frac{\mu_0}{4*\pi} \frac{e* (0 cm, 0cm, -2.0\times 10^5 m^2/s) }{1cm}$

Where to completely determinate the field we should know the material in the given region, here we assumed that is void, so we use mu zero, but this value depends on the material.

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