Question

A 4.05 g bullet moves with a speed of 150 m/s perpendicular to the earth's magnetic field of 5.00×10⁻⁵T . If the bullet possesses a net charge of 1.02×10⁻⁸C , by what distance will it be deflected from its path due to the earth's magnetic field after it has traveled 1.15 km ?

Answer 1

The bullet will be deflected by a distance of 0.01125 m (or 11.25 mm) from its path due to the earth's magnetic field.

Step-by-step explanation:

To calculate the distance the bullet will be deflected from its path due to the earth's magnetic field, we can use the formula for the radius of the circular path an object follows in a magnetic field. The formula is given as:

r = (m * v) / (q * B)

Where:

• r is the radius of the circular path
• m is the mass of the bullet
• v is the velocity of the bullet
• q is the net charge of the bullet
• B is the magnetic field strength

Plugging in the given values:

r = (0.00405 kg * 150 m/s) / (1.02*10^-8 C * 5.00*10^-5 T) = 0.01125 m.

Therefore, the bullet will be deflected by a distance of 0.01125 m (or 11.25 mm) from its path due to the earth's magnetic field.