Answer :
The force \( F \) on a current-carrying wire in a magnetic field is given by the equation:
\[ F = I \cdot L \cdot B \cdot \sin(\theta) \]
where:
- \( I \) is the current in the wire (in amperes, A)
- \( L \) is the length of the wire (in meters, m)
- \( B \) is the magnetic field strength (in teslas, T)
- \( \theta \) is the angle between the current and the magnetic field
In this problem:
- \( I = 4.0 \, \text{A} \)
- \( L = 3.0 \, \text{mm} = 0.003 \, \text{m} \) (since 1 mm = 0.001 m)
- \( B = 0.020 \, \text{T} \)
- \( \theta = 90^\circ \) (since the current is perpendicular to the field, \( \sin(90^\circ) = 1 \))
Substituting these values into the formula:
\[ F = 4.0 \, \text{A} \times 0.003 \, \text{m} \times 0.020 \, \text{T} \times 1 \]
\[ F = 0.00024 \, \text{N} \]
Therefore, the force on the wire is \( 0.00024 \, \text{N} \).
\[ F = I \cdot L \cdot B \cdot \sin(\theta) \]
where:
- \( I \) is the current in the wire (in amperes, A)
- \( L \) is the length of the wire (in meters, m)
- \( B \) is the magnetic field strength (in teslas, T)
- \( \theta \) is the angle between the current and the magnetic field
In this problem:
- \( I = 4.0 \, \text{A} \)
- \( L = 3.0 \, \text{mm} = 0.003 \, \text{m} \) (since 1 mm = 0.001 m)
- \( B = 0.020 \, \text{T} \)
- \( \theta = 90^\circ \) (since the current is perpendicular to the field, \( \sin(90^\circ) = 1 \))
Substituting these values into the formula:
\[ F = 4.0 \, \text{A} \times 0.003 \, \text{m} \times 0.020 \, \text{T} \times 1 \]
\[ F = 0.00024 \, \text{N} \]
Therefore, the force on the wire is \( 0.00024 \, \text{N} \).
