A proton moves along the x-axis with vx=1.0×107m/s.

Asked by maham237 @ 22/12/2020 in Mathematics viewed by 523 People

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)?

Answered by maham237 @ 22/12/2020

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 $0.4\times10^{-15}\ T$ .

A proton moves along the x-axis with vx=1.0×107m/s.

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