Question

In the product $\overrightarrow{\mathbf{F}}\mathbf{=}\mathit{q}\overrightarrow{\mathbf{v}}\mathbf{\times }\overrightarrow{\mathbf{B}}$, take$\mathit{q}\mathbf{=}\mathbf{2}\mathbf{,}\overrightarrow{\mathbf{v}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\stackrel{\mathbf{⏜}}{\mathbf{i}}\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{0}\stackrel{\mathbf{⏜}}{\mathbf{j}}\mathbf{+}\mathbf{60}\stackrel{\mathbf{⏜}}{\mathbf{k}}$,and$\overrightarrow{\mathbf{F}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\stackrel{\mathbf{⏜}}{\mathbf{i}}\mathbf{-}\mathbf{20}\stackrel{\mathbf{⏜}}{\mathbf{j}}\mathbf{+}\mathbf{12}\stackrel{\mathbf{⏜}}{\mathbf{k}}$. What then is$\overrightarrow{\mathbf{B}}$in unit-vector notation if${\mathit{B}}_{\mathbf{x}}\mathbf{=}{\mathit{B}}_{\mathbf{y}}$?

Short answer

If $B_x=B_y$ the vector$\overrightarrow{B}$ can be written as,$-3.0\stackrel{⏜}{i}-3.0\stackrel{⏜}{j}-4.0\stackrel{⏜}{k}$

Step-by-step solution

Given data

$$\begin{gathered} q=2 \\ \overrightarrow{v}=2.0\stackrel{⏜}{i}+4.0\stackrel{⏜}{j}+6.0\stackrel{⏜}{k} \\ \overrightarrow{F}=4.0\stackrel{⏜}{i}-20\stackrel{⏜}{j}+12\stackrel{⏜}{k} \end{gathered}$$

Understanding the concept

Using the formula for cross-product we can find the vector $\overrightarrow{\mathbf{B}}$from the given vectors$\overrightarrow{\mathbf{F}}$and$\overrightarrow{\mathbf{v}}$.

The vector product is written as,

$\left(\overrightarrow{a}\times \stackrel{\rightharpoonup }{b}\right)=\left(a_y{b}_z-b_y{a}_z\right)\stackrel{⏜}{i}+\left(a_z{b}_x-b_z{a}_x\right)\stackrel{⏜}{j}+\left(a_x{b}_y-b_y{a}_x\right)\stackrel{⏜}{k}$ (i)

Calculate B→ in unit-vector notation

Use equation (i) to write the cross product between $\overrightarrow{v}$and $\overrightarrow{B}$.

$$\begin{gathered} \overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B} \\ {F}_x\stackrel{⏜}{i}+F_y\stackrel{⏜}{j}+F_z\stackrel{⏜}{k}=q\left(v_y{B}_z-v_z{B}_y\right)\stackrel{⏜}{i}+q\left(v_z{B}_x-v_x{B}_z\right)\stackrel{⏜}{j}+q\left(v_x{B}_y-v_y{B}_x\right)\stackrel{⏜}{k} \end{gathered}$$

Comparing the above equation with the given vector$\overrightarrow{F}$, we get

$$\begin{gathered} 2\left(4B_z-6B_y\right)=4 \\ 2\left(6B_x-2B_z\right)=-2 \\ 2\left(2B_y-4B_x\right)=12 \end{gathered}$$

As $B_x=B_y$, above equation becomes,

$$\begin{gathered} 2\left(2B_y-4B_y\right)=12 \\ 4B_y-8B_y=12 \\ -4B_y=12 \\ {B}_y=-3 \end{gathered}$$

Inserting this value in above equation we get,

$$\begin{gathered} 2\left(4B_z-6\left(-3\right)\right)=4 \\ {B}_z=-4 \end{gathered}$$

Calculate the value of $B_x$from the above equations.

$$\begin{gathered} B_x=B_y \\ {B}_x=-3 \end{gathered}$$

Therefore, the vector$\overrightarrow{\mathrm{B}}$ can be written as$-3.0\stackrel{⏜}{i}-3.0\stackrel{⏜}{j}-4.0\stackrel{⏜}{k}$