To find the magnitude and direction of the resultant magnetic force on the conducting ring, you can use the formula for the magnetic force on a current-carrying loop:
\[F = B \cdot I \cdot A \cdot \sin(\theta)\]
Where:
- \(F\) is the magnetic force.
- \(B\) is the magnetic field strength.
- \(I\) is the current in the loop.
- \(A\) is the area of the loop.
- \(\theta\) is the angle between the magnetic field direction and the normal to the loop.
Given the values:
- \(B = 3 \, \text{T}\)
- \(I = 3 \, \text{A}\)
- \(A\) (area of the loop) can be calculated as the area of a circle with a radius of 0.10 m (10 cm):
\[A = \pi r^2 = \pi (0.10 \, \text{m})^2\]
- \(\theta = 30^\circ\)
Now, calculate \(A\) and \(\sin(\theta)\):
\[A \approx 0.0314 \, \text{m}^2\]
\(\sin(30^\circ) = 0.5\)
Now, plug these values into the formula:
\[F = (3 \, \text{T}) \cdot (3 \, \text{A}) \cdot (0.0314 \, \text{m}^2) \cdot 0.5\]
Calculate \(F\):
\[F \approx 0.0707 \, \text{N}\]
So, the magnitude of the resultant magnetic force on the ring is approximately \(0.0707 \, \text{N}\).
Now, for the direction, the force is perpendicular to both the magnetic field direction and the current direction, following the right-hand rule. Since the magnetic field is vertical (making an angle of 30 degrees with the vertical), and the current is clockwise, the magnetic force will be directed outward from the center of the ring, parallel to the plane of the ring. You can describe this direction as "outward from the center of the ring" or "perpendicular to the plane of the ring."