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A uniform magnetic field exists in a region which forms an equilateral triangle of side $a$. The field is perpendicular to the plane of the triangle. A charge $q$ enters the triangle with speed $v$ along the $ \perp$ bisector of one of the sides and comes out along the $\perp$ bisector of the other side. Intensity of the magnetic field is

$\begin {array} {1 1} (a)\;\large\frac{mv}{qa} & \quad (b)\;\large\frac{2mv}{qa} \\ (c)\;\large\frac{mv}{2qa} & \quad (d)\;\large\frac{4mv}{qa} \end {array}$


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By using the basic concept of the geometry of the equilateral triangle, one can say that the particle
moves in a circle of radius $ \large\frac{a}{2}$
$B = \large\frac{mv}{qR}$
Ans : (b)
answered Feb 22, 2014 by thanvigandhi_1

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