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Kelvi

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In the product $\vec{F} = q (\vec{v} \times \vec{B}) = q \vec{v} \times \left( \mathbf{B} \hat{i} + \mathbf{B} \hat{j} + \mathbf{B_0} \hat{k} \right),$ given $q = 1$ and $\vec{v} = 2\hat{i} + 4\hat{j} + 6\hat{k}, \newline$ and force $\vec{F} = 4\hat{i} - 20\hat{j} + 12\hat{k},$ what will be the complete expression for $\vec{B}?$

hard

Moving Charges and Magnetism

2021

physics

$- 8\hat{i} - 8\hat{j} - 6\hat{k}$

$- 6\hat{i} - 6\hat{j} - 8\hat{k}$

Explanation

To solve this problem, we need to find the magnetic field vector $\vec{B}$ such that the force $\vec{F}$

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$8\hat{i} + 8\hat{j} - 6\hat{k}$

$6\hat{i} + 6\hat{j} - 8\hat{k}$

is given by the equation:

\\ \vec{F} = q (\vec{v} \times \vec{B}) \\

Given:

\\ q = 1, \\ \vec{v} = 2\hat{i} + 4\hat{j} + 6\hat{k}, \\ \vec{F} = 4\hat{i} - 20\hat{j} + 12\hat{k}, \\ \vec{B} = B_x \hat{i} + B_y \hat{j} + B_0 \hat{k} \\

We need to compute the cross product

\vec{v} \times \vec{B}: \\ \vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & 6 \\ B_x & B_y & B_0 \\ \end{vmatrix} \\

Expanding the determinant, we get:

\\ \vec{v} \times \vec{B} = \hat{i}(4B_0 - 6B_y) - \hat{j}(2B_0 - 6B_x) + \hat{k}(2B_y - 4B_x) \\

Equating this to the given force

\vec{F} = 4\hat{i} - 20\hat{j} + 12\hat{k},

we have:

\\ 4B_0 - 6B_y = 4 \quad

(1)

\\ 2B_0 - 6B_x = 20 \quad

(2)

\\ 2B_y - 4B_x = 12 \quad

(3)

\\

Solving these equations:

\\

From equation (1):

\\ 4B_0 - 6B_y = 4 \\ B_0 = \frac{4 + 6B_y}{4} \\ B_0 = 1.5 + 1.5B_y \quad

(4)

\\

From equation (2):

\\ 2B_0 - 6B_x = 20 \\ B_0 = 10 + 3B_x \quad

(5)

\\

Equating equations (4) and (5):

\\ 1.5 + 1.5B_y = 10 + 3B_x \\ 1.5B_y = 8.5 + 3B_x \\ B_y = \frac{8.5 + 3B_x}{1.5} \\ B_y = \frac{17 + 6B_x}{3} \quad

(6)

\\

Substitute

B_y

from equation (6) into equation (3):

\\ 2\left(\frac{17 + 6B_x}{3}\right) - 4B_x = 12 \\ \frac{34 + 12B_x}{3} - 4B_x = 12 \\ 34 + 12B_x - 12B_x = 36 \\ 34 = 36 \quad

(This is incorrect, indicating a calculation error)

\\

Re-evaluate the equations:

\\

From equation (1):

\\ 4B_0 - 6B_y = 4 \\ B_0 = 1 + 1.5B_y \\

From equation (2):

\\ 2B_0 - 6B_x = 20 \\ B_0 = 10 + 3B_x \\

Equate the two expressions for

B_0: \\ 1 + 1.5B_y = 10 + 3B_x \\ 1.5B_y = 9 + 3B_x \\ B_y = 6 + 2B_x \\

Substitute

B_y = 6 + 2B_x

into equation (3):

\\ 2(6 + 2B_x) - 4B_x = 12 \\ 12 + 4B_x - 4B_x = 12 \\ 12 = 12 \quad

(This is consistent)

\\

Now, solve for

B_x: \\ B_y = 6 + 2B_x \\ B_0 = 10 + 3B_x \\

Using the options, substitute and verify:

\\

Option 2:

-6\hat{i} - 6\hat{j} - 8\hat{k} \\ B_x = -6, \\ B_y = -6, \\ B_0 = -8 \\

Check:

\\ B_y = 6 + 2(-6) = 6 - 12 = -6 \\ B_0 = 10 + 3(-6) = 10 - 18 = -8 \\

This satisfies all equations, so the correct option is Option 2.