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In the product F⃗=q(v⃗×B⃗)=qv⃗×(Bi^+Bj^+B0k^)\vec{F} = q(\vec{v} \times \vec{B}) = q\vec{v} \times (B\hat{i} + B\hat{j} + B_0\hat{k}). For q=1q=1 and v⃗=2i^+4j^+6k^\vec{v} = 2\hat{i} + 4\hat{j} + 6\hat{k} and F⃗=4i^−20j^+12k^\vec{F} = 4\hat{i} - 20\hat{j} + 12\hat{k}, what will be the complete expression for B⃗\vec{B}?

A

8i^+8j^−6k^8\hat{i} + 8\hat{j} - 6\hat{k}

B

6i^+6j^−8k^6\hat{i} + 6\hat{j} - 8\hat{k}

C

−8i^−8j^−6k^-8\hat{i} - 8\hat{j} - 6\hat{k}

D

−6i^−6j^−8k^-6\hat{i} - 6\hat{j} - 8\hat{k}

Step-by-Step Solution

  1. Identify Formula: The force on a moving charge is given by the Lorentz force equation F⃗=q(v⃗×B⃗)\vec{F} = q(\vec{v} \times \vec{B}).
  2. Substitute Knowns: q=1q = 1 v⃗=2i^+4j^+6k^\vec{v} = 2\hat{i} + 4\hat{j} + 6\hat{k} F⃗=4i^−20j^+12k^\vec{F} = 4\hat{i} - 20\hat{j} + 12\hat{k} B⃗\vec{B} is assumed to be of the form Bxi^+Byj^+Bzk^B_x\hat{i} + B_y\hat{j} + B_z\hat{k}. The problem states B⃗=Bi^+Bj^+B0k^\vec{B} = B\hat{i} + B\hat{j} + B_0\hat{k}, implying Bx=ByB_x = B_y.
  3. Evaluate Options: We can test the options by calculating the cross product v⃗×B⃗\vec{v} \times \vec{B} for each.
  • Test Option D: B⃗=−6i^−6j^−8k^\vec{B} = -6\hat{i} - 6\hat{j} - 8\hat{k} v⃗×B⃗=∣i^j^k^246−6−6−8∣\vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\2 & 4 & 6 \\ -6 & -6 & -8 \end{vmatrix} i^\hat{i}-component: (4)(−8)−(6)(−6)=−32+36=4(4)(-8) - (6)(-6) = -32 + 36 = 4 j^\hat{j}-component: −[(2)(−8)−(6)(−6)]=−[−16+36]=−20-[(2)(-8) - (6)(-6)] = -[-16 + 36] = -20 k^\hat{k}-component: (2)(−6)−(4)(−6)=−12+24=12(2)(-6) - (4)(-6) = -12 + 24 = 12 Result: 4i^−20j^+12k^4\hat{i} - 20\hat{j} + 12\hat{k}.
  1. Conclusion: The calculated force matches the given force F⃗\vec{F}. Therefore, the magnetic field vector is B⃗=−6i^−6j^−8k^\vec{B} = -6\hat{i} - 6\hat{j} - 8\hat{k}.
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