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JEE Main 2017 · 8 Apr · Shift 1, Physics question 18

In a certain region static electric and magnetic fields exist. The magnetic field is given by B⃗=B0(i^+2j^−4k^)\vec B=B_0\left(\hat i+2\hat j-4\hat k\right). If a test charge moving with a velocity v⃗=v0(3i^−j^+2k^)\vec v=v_0\left(3\hat i-\hat j+2\hat k\right) experiences no force in that region, then the electric field in the region, in SI units, is :

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Answer and worked solution

Official answer: D

No force means E⃗=−v⃗×B⃗\vec E=-\vec v\times\vec B.
v⃗×B⃗=v0B0∣i^j^k^3−1212−4∣=v0B0(0i^+14j^+7k^).\vec v\times\vec B=v_0B_0\begin{vmatrix}\hat i&\hat j&\hat k\\3&-1&2\\1&2&-4\end{vmatrix}=v_0B_0(0\hat i+14\hat j+7\hat k).
So E⃗=−v0B0(14j^+7k^)\vec E=-v_0B_0(14\hat j+7\hat k).

Topics:Motion in magnetic and electric fieldsCross product
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