To find the magnitude of the component of the magnetic field that is perpendicular to the wire, we can use the equation for the magnetic force exerted on a current-carrying conductor in a magnetic field. The equation is given by:
\[ F = B \cdot I \cdot L \cdot \sin(\theta) \]
where:
- \( F \) is the magnetic force
- \( B \) is the magnetic field strength
- \( I \) is the current
- \( L \) is the length of the conductor
- \( \theta \) is the angle between the direction of the current and the direction of the magnetic field
According to the question, the magnetic force (\( F \)) on the wire is 1.0 N, the length of the wire segment (\( L \)) is 0.15 m, and the current (\( I \)) is 4.5 A. Since the magnetic force is perpendicular to the wire, this implies that \( \theta \) is 90 degrees. The sine of 90 degrees is 1.
So we can plug the values we have into the equation to solve for \( B \):
\[ 1.0 = B \cdot 4.5 \cdot 0.15 \cdot \sin(90^\circ) \]
As \( \sin(90^\circ) = 1 \), the equation simplifies to:
\[ 1.0 = B \cdot 4.5 \cdot 0.15 \]
Now we can solve for the magnetic field (\( B \)):
\[ B = \frac{1.0}{(4.5 \cdot 0.15)} \]
Let's do the calculation:
\[ B = \frac{1.0}{0.675} \]
\[ B \approx 1.48148 \]
\[ B \approx 1.48 \] (rounded to two decimal places)
Hence, the magnitude of the component of the magnetic field that is perpendicular to the wire is approximately 1.48 Tesla.
