Question

A uniform magnetic field exists in a region which forms an equilateral triangle of side $a$. The magnetic field is perpendicular to the plane of the triangle. A charge $q$ of enters into this magnetic field perpendicular to a side with speed $v$. The charge enters from midpoint and leaves the field from mid-point of other side. Magnetic induction in the triangle is (Given mass of charge $=m$)

• A. $\frac{mv}{qa}$
• B. $\frac{2mv}{qa}$
• C. $\frac{mv}{2qa}$
• D. $\frac{mv}{4qa}$

Answer 1

Answer: (B)

$\frac{2mv}{qa}$

As shown in the figure, the motion is symmetric about the entering and leaving sides.

The radius of the circle is $r=\frac{a}{2}$ where a is the side length of the equilateral triangle.
$r=\frac{mv}{qB}$

$\therefore B=\frac{mv}{qr}=\frac{mv}{q\frac{a}{2}}=\frac{2mv}{qa}$