Find the direction and magnitude of the minimum velocity of charge q = +2e that will produce the magnetic force F shown in the figure below. (Measure the angle counterclockwise from the x axis in the xz plane. Take the magnitudes of F as 2.11 10-11 N and the magnitude of B as 1.83 T.) magnitude direction

Find the direction and magnitude of the minimum velocity of charge q = +2e that will produce the magnetic force F shown in the figure below. (Measure the angle counterclockwise from the x axis in the xz plane. Take the magnitudes of F as 2.11 10-11 N and the magnitude of B as 1.83 T.) magnitude direction

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Question

magnitude
direction

Expert Solution

Step 1

Given,

charge of the particle is

$\begin{array}{rcl} q & = & + 2 e \\ = & 2 \times 1.6 \times 10^{- 19} (C) \\ = & 3.2 \times 10^{- 19} (C) \end{array}$

the magnitude of the magnetic field is $B = 1.83 (T)$

And the magnetic field is in the y direction.

Then,

$\vec{B} = 1.83 \hat{j} (T)$

And the magnitude of the force is $F = 2.11 \times 10^{- 11} (N)$

The force is in xz plane and it is 45^o from the x-axis.

Then the x component of force is

$\begin{array}{rcl} F_{x} & = & F \cos (45^{o}) \\ = & 2.11 \times 10^{- 11} (N) \times \cos (45^{o}) \\ = & 1.5 \times 10^{- 11} (N) \end{array}$

the y component of force is

$\begin{array}{rcl} F_{y} & = & 0 \end{array}$

the z component of force is

$\begin{array}{rcl} F_{z} & = & F \cos (45^{o}) \\ = & 2.11 \times 10^{- 11} (N) \times \sin (45^{o}) \\ = & 1.5 \times 10^{- 11} (N) \end{array}$

So force will be

$\begin{array}{rcl} \vec{F} & = & 1.5 \times 10^{- 11} (N) \hat{i} + 1.5 \times 10^{- 11} (N) \hat{k} \\ = & 1.5 \times 10^{- 11} (\hat{i} + \hat{k}) (N) \end{array}$

Let the velocity of the charged particle is

$\vec{v} = v_{x} \hat{i} + v_{y} \hat{j} + v_{z} \hat{k}$

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