An electron is projected with velocity `v_(0)` in a unifrom electric field E perpendicular to the field Again it is projectced with velocity `v_(0)` perpendicular to a unifrom magnetic field `B .` if `r_(1)` is initial radius of curvature just after entering in the electric field and `r_(2)` is initial radius of curvature just after entering in magnetic field then the ratio `r_(1)//r_(2)` is equal to .

Options

A. `(Bv(2)o)/(E)`
B. `(B)/(E)`
C. `(Evo)/(B)`
D. `(Bvo)/(E)`

Answer

To solve the problem, we need to find the ratio of the initial radius of curvature of an electron projected into a uniform electric field \(E\) and a uniform magnetic field \(B\). Let's denote:

- \(r_1\) as the radius of curvature in the electric field,
- \(r_2\) as the radius of curvature in the magnetic field,
- \(v_0\) as the initial velocity of the electron,
- \(q\) as the charge of the electron,
- \(m\) as the mass of the electron.

### Step-by-Step Solution

**Step 1: Determine the radius of curvature in the electric field (\(r_1\))**

In a uniform electric field, the force acting on the electron is given by:
\[ F_E = qE \]
This force provides the centripetal force required for circular motion. The centripetal force can be expressed as:
\[ F_c = \frac{mv_0^2}{r_1} \]
Setting the two forces equal gives:
\[ qE = \frac{mv_0^2}{r_1} \]
Rearranging this equation to solve for \(r_1\):
\[ r_1 = \frac{mv_0^2}{qE} \]

**Step 2: Determine the radius of curvature in the magnetic field (\(r_2\))**

In a uniform magnetic field, the magnetic force acting on the electron is given by:
\[ F_B = qv_0B \]
Again, this force provides the centripetal force for circular motion:
\[ F_c = \frac{mv_0^2}{r_2} \]
Setting the forces equal gives:
\[ qv_0B = \frac{mv_0^2}{r_2} \]
Rearranging this equation to solve for \(r_2\):
\[ r_2 = \frac{mv_0}{qB} \]

**Step 3: Find the ratio \( \frac{r_1}{r_2} \)**

Now we can find the ratio of the two radii of curvature:
\[ \frac{r_1}{r_2} = \frac{\frac{mv_0^2}{qE}}{\frac{mv_0}{qB}} \]
This simplifies to:
\[ \frac{r_1}{r_2} = \frac{mv_0^2}{qE} \cdot \frac{qB}{mv_0} = \frac{v_0B}{E} \]

### Final Result

Thus, the ratio of the initial radius of curvature just after entering the electric field to that just after entering the magnetic field is:
\[ \frac{r_1}{r_2} = \frac{v_0B}{E} \]