A charged particle of charge `2C` thrown vertically upwards `10 m//s`.Find the magnetic force on this charge due to earth's magnetic field.Given vertical component of the earth `=3muT` and angle of dip `=37^(@)`

Explanation

To solve the problem of finding the magnetic force on a charged particle thrown vertically upwards, we can follow these steps:

### Step 1: Understand the Given Data
- Charge of the particle, \( Q = 2 \, C \)
- Initial velocity of the particle, \( V = 10 \, m/s \)
- Vertical component of Earth's magnetic field, \( B_v = 3 \, \mu T = 3 \times 10^{-6} \, T \)
- Angle of dip, \( \theta = 37^\circ \)

### Step 2: Find the Horizontal Component of Earth's Magnetic Field
Using the angle of dip, we can relate the vertical and horizontal components of the Earth's magnetic field. The tangent of the angle of dip is given by:

\[ \tan(\theta) = \frac{B_v}{B_h} \]

Where:
- \( B_h \) is the horizontal component of the magnetic field.

From the given angle of dip \( \theta = 37^\circ \):

\[ \tan(37^\circ) = \frac{3}{4} \]

This implies:

\[ \frac{B_v}{B_h} = \frac{3}{4} \]

Rearranging gives:

\[ B_h = \frac{4}{3} B_v \]

Substituting \( B_v \):

\[ B_h = \frac{4}{3} \times 3 \times 10^{-6} \, T = 4 \times 10^{-6} \, T \]

### Step 3: Calculate the Magnetic Force
The magnetic force \( F \) on a charged particle moving in a magnetic field is given by the formula:

\[ F = Q \cdot V \cdot B_h \]

Substituting the known values:

\[ F = 2 \, C \cdot 10 \, m/s \cdot 4 \times 10^{-6} \, T \]

Calculating this gives:

\[ F = 2 \cdot 10 \cdot 4 \times 10^{-6} = 80 \times 10^{-6} \, N = 8 \times 10^{-5} \, N \]

### Final Answer
The magnetic force on the charged particle is:

\[ F = 8 \times 10^{-5} \, N \]

---