A charge q is moving with a velocity $v_{1} = 1 \hat{i} m / s$ at a point in a magnetic field and experience a force $F = q [- \hat{j} + 1 \hat{k}] N .$ If the charge is moving with a velocity $v_{2} = 1 \hat{j} m / s$ at the same point, it experiences a force $F_{2} = q (1 \hat{i} - 1 \hat{k}) N$ . The magnetic induction B at that point is

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$(\hat{i} - \hat{j} + \hat{k}) Wb / m^{2}$

$(\hat{i} + \hat{j} - \hat{k}) Wb / m^{2}$

$(- \hat{i} + \hat{j} - \hat{k}) Wb / m^{2}$

$(\hat{i} + \hat{j} + \hat{k}) Wb / m^{2}$

answer is A.

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Detailed Solution

Let the magnetic field is
$B = B_{1} \hat{i} + B_{2} \hat{j} + B_{3} \hat{k}$
Applying, $F_{m} = q (v \times B)$ two times, we have
$q [- \hat{j} + \hat{k}] =$

$q [(\hat{i}) \times (B_{1} \hat{i} + B_{2} \hat{j} + B_{3} \hat{k})] = q [B_{2} \hat{k} - B_{3} \hat{j}]$
Comparing two sides, we get,
B₂=1 and B₃=1
Further,
$q [\hat{i} - \hat{k}] =$

$q [(\hat{j}) \times (B_{1} \hat{i} + B_{2} \hat{j} + B_{3} \hat{k})] = q [- B_{1} \hat{k} + B_{3} \hat{i}]$
Again comparing, we get
$\begin{array}{l} ∴ B_{1} = 1 and B_{3} = 1 \\ B = (\hat{i} + \hat{j} + \hat{k}) Wb / m^{2} \end{array}$