Question

A point charge $q = 2 μC$ is at the origin. It has velocity $2 \hat{i} m s^{- 1}$ . Find the magnetic field at the following points in vector form (at the moment when the charged particle passes through the origin):

(i) $(2, 0, 0)$

(ii) $(0, 2, 0)$

(iii) $(0, 0, 2)$

(iv) $(2, 1, 2)$

(v) Is the magnitude of the magnetic field on the circumference of the circle (in $y z$ plane) $y^{2} + z^{2} = c^{2}$ , where $c$ is a constant is same everywhere. Is it same in direction also?

(vi) Answer the above (v) for the circle of same equation but in a plane $x = a$ , where $a$ is a constant.

Accepted Answer

Here $q = 2 μ C$ at origin with velocity $\vec{v} = 2 \hat{i} m s^{- 1}$

$(i) (2, 0, 0)$

By using Biot-Savart law in vector form;

$B at (2, 0, 0) = \frac{μ_{0}}{4 π} q \frac{(\vec{v} \times \vec{r})}{r^{3}}$

$B at (2, 0, 0) = 0; as \vec{v} ∥ \vec{r} \vec{r} = 2 \hat{i}$

$(ii) B at (0, 2, 0) = \frac{μ_{0}}{4 π} q \frac{(\vec{v} \times \vec{r})}{r^{3}}; \vec{v} = 2 \hat{i} m s^{- 1} \vec{r} = 2 \hat{j} B at (0, 2, 0) = \frac{μ_{0} q \hat{k}}{8 π}$

$(iii) Similarly for (0, 0, 2) \vec{r} = 2 \hat{k} \vec{v} = 2 \hat{i} m s^{- 1} B at (0, 0, 2) = - \frac{μ_{0} q (4) \hat{j}}{(4 π) 8} = - \frac{μ_{0} q \hat{j}}{8 π}$

$(iv) B at (2, 1, 2) \vec{r} = 2 \hat{i} + \hat{j} + 2 \hat{k}$

$\vec{B} = \frac{μ_{0}}{4 π} q [\frac{2 \hat{i} \times (2 \hat{i} + \hat{j} + 2 \hat{k})}{{(3)}^{3}}]$

$= \frac{μ_{0}}{4 π} q [\frac{(2 \hat{k} + 4 \hat{j})}{27}]$

$(v) At point on the circumference of the circle, the coordinates are \vec{r} = r (cosθ \hat{j} + sinθ \hat{k})$

$B at circumference = \frac{μ_{0}}{4 π} q \frac{(\vec{v} \times \vec{r})}{r^{3}}$

$= (\frac{μ_{0}}{4 π} q) (\frac{2 \hat{i} \times r (\cos θ \hat{j} + \sin θ \hat{k})}{r^{3}})$

$= (\frac{μ_{0} q}{2 π}) (\frac{(\cos θ \hat{k} - \sin θ \hat{j})}{r^{2}})$

Magnitude of magnetic field $|\vec{B}|$ is same on the circumference of the circle, but direction will be not same on the circumference.

$(vi) For this part {(x - a)}^{2} + y^{2} + z^{2} = C^{2}$

$At point on the circumference of the circle, the coordinates are \vec{r} = a \hat{i} + R (cosθ \hat{j} + sinθ \hat{k})$

$B at circumference = \frac{μ_{0}}{4 π} q \frac{(\vec{v} \times \vec{r})}{r^{3}}$

$= (\frac{μ_{0}}{4 π} q) (\frac{2 \hat{i} \times [a \hat{i} + R (cosθ \hat{j} + sinθ \hat{k})]}{{(\sqrt{a^{2} + R^{2}})}^{3}})$

$= (\frac{μ_{0} q}{2 π}) (\frac{R (\cos θ \hat{k} - \sin θ \hat{j})}{{(\sqrt{a^{2} + R^{2}})}^{3}})$

Magnitude of magnetic field $|\vec{B}|$ is same on the circumference of the circle, but direction will be not same on the circumference.

Important Questions on Magnetic Effect of Current

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