A particle with charge 2.0 C moves through a uniform magnetic field. At one instant the velocity of the particle is $\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{+}\mathbf{6}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}\mathbf{)}\mathbf{}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}\mathbf{}$and the magnetic force on the particle is $\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{-}\mathbf{20}\hat{\mathbf{j}}\mathbf{}\mathbf{+}\mathbf{12}\hat{\mathbf{k}}\mathbf{}\mathbf{}\mathbf{)}\mathit{N}\mathbf{}$The xand ycomponents of the magnetic field are equal. What is $\overrightarrow{B}$?

1. The charge on the particle is,$q=2.0C$.
2. The velocity of the particle is,$v⃗=(2.0\hat{i}+4.0\hat{j}+6.0\hat{k})m/s$
3. The force acting on the particle is,$\left(F_B\right)⃗=(4.0\hat{i}-20\hat{j}+12\hat{k})N$.

By using the formula for magnetic force acting on a moving charged particle, we can find the magnetic field B.

Formula:

Magnetic force,$\mathbf{\left(}{\mathit{F}}_{\mathbf{B}}\mathbf{\right)}\mathbf{⃗}\mathbf{=}\mathit{q}\mathit{v}\mathbf{⃗}\mathbf{\times }\mathit{B}\mathbf{⃗}$

To findthemagnetic field, we usetheformula for magnetic force in terms of velocity and magnetic field.

Let the magnetic field vector be $B⃗=(B_x\hat{i}+B_y\hat{j}+B_z\hat{k})T$

As it is given that x and y components of magnetic field are same, B_x=B_y

$B⃗=(B_x\hat{i}+B_y\hat{j}+B_z\hat{k})T$

We have

$\left(F_B\right)⃗=qv⃗\times B⃗$

Therefore, we can write

role="math" localid="1662876658877" $4.0\hat{i}-20\hat{j}+12\hat{k}=2\times \left(\begin{array}{ccc}i& j& k\\ 2& 4& 6\\ B_x& B_x& B_z\end{array}\right)$

By solving the matrix,

$$\begin{gathered} \frac{1}{2}\left(4.0\hat{i}-20\hat{j}+12\hat{k}\right)=i\left(4B_z-6B_x\right)-j(2B_z-6B_x)+k(2B_x-4B_x) \\ \left(2.0\hat{i}-10\hat{j}+6\hat{k}\right)=i\left(4B_z-6B_x\right)-j(2B_z-6B_x)-k\left(2B_x\right) \end{gathered}$$

Comparing both sides, we get

$2=4B_z-6B_x$ …(1)

$10=2B_z-6B_x$

$6=-2B_x$ …(2)

From equation 2,B_x=-3

Using in equation 1,

role="math" localid="1662877154991" $\begin{array}{l}2=4B_z-6\times \left(-3\right)\\ B_z=\frac{16}{4}\\ B_z=-4\end{array}$

Hence x,y,z components ofthemagnetic field will be
$B_x=-3,B_y=-3,B_z=-4$

$B⃗=(-3.0\hat{i}-3.0\hat{j}-4.0\hat{k})T$