A uniform magnetic field exist in a region which forms an equilateral triangle of side a. The magnetic field is perpendicular to the plane of the triangle . A charged q enters into this magnetic field perpendicular to a side with speed v. The charge enters from midpoint and leaves the field from midpoint of other side. Magnetic induction in the triangles is

Options

A. `(mv)/(qa)`
B. `(2mv)/(qa)`
C. `(mv)/(2qa)`
D. `(mv)/(4qa)`

Answer

To solve the problem, we need to analyze the motion of a charged particle in a magnetic field within an equilateral triangle. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Geometry of the Problem
We have an equilateral triangle with side length \( a \). A charged particle \( q \) enters the magnetic field from the midpoint of one side (let's call it side \( PQ \)) and exits from the midpoint of the opposite side (side \( QR \)). The magnetic field \( B \) is perpendicular to the plane of the triangle.

**Hint:** Visualize the triangle and mark the midpoints where the charge enters and exits.

### Step 2: Determine the Radius of Circular Motion
Since the charge enters the magnetic field perpendicularly and follows a circular path, we can determine the radius \( R \) of the circular motion. The distance from the midpoint of side \( PQ \) to the midpoint of side \( QR \) is the height of the triangle. The height \( h \) of an equilateral triangle can be calculated as:
\[ h = \frac{\sqrt{3}}{2} a \]
However, the radius \( R \) of the circular path taken by the charge is half the height, which is:
\[ R = \frac{h}{2} = \frac{\sqrt{3}}{4} a \]

**Hint:** Remember that the radius of the circular path is related to the geometry of the triangle.

### Step 3: Apply the Lorentz Force
The force acting on the charged particle due to the magnetic field is given by the Lorentz force:
\[ F = qvB \]
This force acts as the centripetal force required to keep the charge moving in a circular path:
\[ F = \frac{mv^2}{R} \]

**Hint:** Identify the relationship between the Lorentz force and centripetal force.

### Step 4: Equate the Forces
Setting the two expressions for force equal gives:
\[ qvB = \frac{mv^2}{R} \]
From this, we can solve for the magnetic field \( B \):
\[ B = \frac{mv}{qR} \]

**Hint:** Rearranging the equation helps isolate the variable of interest.

### Step 5: Substitute the Radius
Now, substitute \( R = \frac{\sqrt{3}}{4} a \) into the equation for \( B \):
\[ B = \frac{mv}{q \left(\frac{\sqrt{3}}{4} a\right)} = \frac{4mv}{q\sqrt{3} a} \]

**Hint:** Ensure that you substitute the radius correctly to find the magnetic field.

### Final Answer
Thus, the magnetic induction (magnetic field) in the triangle is:
\[ B = \frac{4mv}{q\sqrt{3} a} \]