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Electric Field Due to a Dipole Along the Equatorial Line

When a partic...

Question

When a particle of charge q is projected with uniform speed u along x - axis of the Cartesian coordinate system $(^i,^j,^k)$ in the presence of a magnetic field of induction $\to B$ , the force on q is $\to F = \frac{q u \to B}{2}^j$ and when the particle is projected along y - axis with same speed, the force $\to F = \frac{q u \to B}{2} (-^i + \sqrt{3}^k)$ . The magnetic field $\to B$ is

A

\frac{B}{2} (\sqrt{3}^i +^k)

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B

\frac{B}{2} (\sqrt{3}^i -^k)

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C

- \frac{B}{2} (\sqrt{3}^i +^k)

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D

\frac{B}{2} (- \sqrt{3}^i +^k)

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Solution

The correct option is B $- \frac{B}{2} (\sqrt{3}^i +^k)$
Initial velocity along x-axis $U = U^i$
Force when projected along x-axis $F = q (\to V \times \to B)$
Let the magnetic field be $\to B = B (a^i + b^j + c^v)$
$V_{1} = U^i$
$F_{1} = \frac{q U B}{2}^j = q B (U^i \times (a^i + b^j + c^k))$
$\frac{q U B}{2}^j = q B u (v^k - c^j)$
$b^k - c^j = \frac{^j}{2}$
$c = - \frac{1}{2}, b = 0$
$F_{2} = \frac{q U \to B}{2} (-^i + \sqrt{3}^k), V_{2} = U^j$
$F_{2} = q (U^j \times \to B (a^i + c^k))$
$\frac{q V \to B}{2} (-^i + \sqrt{3}^k) = q V \to B (-^k a + c^i)$
$a = - \frac{\sqrt{3}}{2}, c = - \frac{1}{2}$
$\to B = - \to B (\frac{\sqrt{3}}{2}^i + \frac{1}{2}^k)$

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Electric Field Due to a Dipole Along the Equatorial Line

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