Question

A uniform magnetic field exists in region which forms an equilateral triangle of side $a$. The magnetic field is perpendicular to the plane of the triangle. A charge $q$ enters into the magnetic field perpendicularly with speed $v$ along perpendicular bisector of one side. The value the magnetic field is ? (Assume that width of magnetic field region is small to complete full circular path).

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

Answer 1

The correct option is C $\frac{2mv}{qa}$

The given scenario is depicted in the figure shown.



Due to symmetry of motion. The particle will also emerge perpendiculary from another side.

From the geometry it is clear that radius of circular path, $R=\frac{a}{2}$

$\Rightarrow \frac{mv}{qB}=\frac{a}{2}$

$\therefore B=\frac{2mv}{qa}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option (c) is the correct answer.

Why this question ? Since it is given that magnetic field region width is not sufficient describing full circle, hence due to motion symmetry the charge will come out along bisector of side lying along its path.