Name: Magnetomotive Force
Uploaded: 2019-07-19
Duration: 2 min 40 s
Description: You might have seen the equation of magnetic flux. There appears a quantity which is known as magnetomotive force. Let's watch the video to learn more.

Question

A positive point charge

q

of mass

m

, kept at a distance

x_{0}

(in the same plane) from a fixed very long straight current is projected normally away from it with speed

v

. Find the maximum separation between the wire and the particle.

Accepted Answer

At minimum separation the $x -$ component of velocity of the particle will be zero. Find the acceleration of the particle due to the magnetic force and solve to get the expression for velocity and displacement. Let the particle be initially at $P$ . Take the wire as the $Y -$ axis and the foot of perpendicular from $P$ to the wire as the origin. Take the line $O P$ as the $X -$ axis. We have, $O P = x_{\circ}$ . The magnetic field $\vec{B}$ at any point to the right of the wire is along the negative $Z -$ axis. The magnetic force on the particle is, therefore, in the $X - Y$ plane. As there is no initial velocity along the $Z -$ axis, the motion will be in the $X - Y$ plane. Also, its speed remains unchanged. As the magnetic field is not uniform, the particle does not go along a circle.

The force at time $t$ is

$\vec{F} = q (\vec{V} \times \vec{B})$

$= q (v_{x} \hat{i} + v_{y} \hat{j}) \times (- \frac{μ_{0} i \hat{k}}{2 π x})$

$= q v_{x} \frac{μ_{0} i}{2 π x} \hat{j} - q v_{y} \frac{μ_{0} i}{2 π x} \hat{j}$

Thus, $a_{x} = \frac{F_{x}}{m} = \frac{- μ_{0} q_{i}}{2 π m} \frac{μ_{y}}{x} = - λ \frac{μ_{y}}{x} . . . (i)$

where $λ = \frac{μ_{0} q i}{2 π m}$

Also, $a_{x} = \frac{d V_{x}}{d t} = \frac{d v x}{d x} \times \frac{d x}{d t} = \frac{v_{x} d v_{x}}{d x} . . . (i i)$

As $v_{x}^{2} + v_{y_{0}}^{2} = v^{2}$

Differentiating both sides

$2 v_{x} d v_{x} + 2 v_{y} d v_{y} = 0$

giving $V_{x} d v_{x} = - V_{y} d y . . . (i i i)$

From $(i), (i i) & (i i i)$

$\frac{V_{y} d y_{y}}{d x} = \frac{λ v_{y}}{x}, \frac{d x}{x} = \frac{d x_{y}}{λ}$

initially $x = x_{0}$ and $V_{y} = 0$ . At minimum separation from the wire $v_{x} = x$ so that $v_{y} = v$

thus $\int_{x_{0}}^{x} \frac{d x}{x} = \int_{0}^{- v} \frac{d v_{y}}{λ}$ or

$\ln \frac{x}{x_{0}} = - \frac{v}{λ}$ or $x = x_{0} e^{- \frac{v}{λ}}$

$= x_{0} e^{\frac{^{- 2 π m v}}{μ_{0} q i}}$

A positive point charge $q$ of mass $m$ , kept at a distance $x_{0}$ (in the same plane) from a fixed very long straight current is projected normally away from it with speed $v$ . Find the maximum separation between the wire and the particle.

Important Questions on Magnetic Effect of Current

Learn and Prepare for any exam you want !

A positive point charge q of mass m, kept at a distance x0 (in the same plane) from a fixed very long straight current is projected normally away from it with speed v. Find the maximum separation between the wire and the particle.

Important Questions on Magnetic Effect of Current

Learn and Prepare for any exam you want !

A positive point charge $q$ of mass $m$ , kept at a distance $x_{0}$ (in the same plane) from a fixed very long straight current is projected normally away from it with speed $v$ . Find the maximum separation between the wire and the particle.