An l-shaped metal machine part is made of two equal-length segments that are perpendicular to each other and carry a 4.50-a current as shown in the figure. This part has a total mass of 3.80 kg and a total length of 3.00 m, and it is in an external 1.20-t magnetic field that is oriented perpendicular to the plane of the part, as shown. What is the magnitude of the net magnetic force that the field exerts on the part?
The Correct answer and Explanation is:
To calculate the net magnetic force exerted on the L-shaped metal machine part by the external magnetic field, we can use the formula for the magnetic force on a current-carrying conductor, which is given by:F=I⋅L⋅B⋅sin(θ)F = I \cdot L \cdot B \cdot \sin(\theta)F=I⋅L⋅B⋅sin(θ)
where:
- FFF is the magnetic force,
- III is the current through the conductor,
- LLL is the length of the conductor within the magnetic field,
- BBB is the magnetic field strength, and
- θ\thetaθ is the angle between the current direction and the magnetic field.
Given Values
- Current, I=4.50 AI = 4.50 \, AI=4.50A
- Total length of the L-shaped part, Ltotal=3.00 mL_{total} = 3.00 \, mLtotal=3.00m (each segment is 1.50 m1.50 \, m1.50m since they are equal)
- Magnetic field strength, B=1.20 TB = 1.20 \, TB=1.20T
- The magnetic field is perpendicular to the plane of the part, implying that the angle θ=90∘\theta = 90^\circθ=90∘, and sin(90∘)=1\sin(90^\circ) = 1sin(90∘)=1.
Calculating Force for Each Segment
The L-shaped part consists of two equal-length segments, each of 1.50 m1.50 \, m1.50m. Therefore, we can calculate the force for each segment:Fsegment=I⋅Lsegment⋅B⋅sin(90∘)=4.50 A⋅1.50 m⋅1.20 T⋅1F_{segment} = I \cdot L_{segment} \cdot B \cdot \sin(90^\circ) = 4.50 \, A \cdot 1.50 \, m \cdot 1.20 \, T \cdot 1Fsegment=I⋅Lsegment⋅B⋅sin(90∘)=4.50A⋅1.50m⋅1.20T⋅1
Calculating this:Fsegment=4.50⋅1.50⋅1.20=8.10 NF_{segment} = 4.50 \cdot 1.50 \cdot 1.20 = 8.10 \, NFsegment=4.50⋅1.50⋅1.20=8.10N
Direction of Forces
The magnetic forces on the two segments will act in perpendicular directions due to the orientation of the current. When the current flows in the segments, the forces exerted will be directed according to the right-hand rule.
Net Magnetic Force
Since the two forces act at right angles to each other, we can use the Pythagorean theorem to find the net force:Fnet=F12+F22=(8.10 N)2+(8.10 N)2=2⋅(8.10 N)2=8.102≈11.44 NF_{net} = \sqrt{F_{1}^2 + F_{2}^2} = \sqrt{(8.10 \, N)^2 + (8.10 \, N)^2} = \sqrt{2 \cdot (8.10 \, N)^2} = 8.10 \sqrt{2} \approx 11.44 \, NFnet=F12+F22=(8.10N)2+(8.10N)2=2⋅(8.10N)2=8.102≈11.44N
Conclusion
Thus, the magnitude of the net magnetic force exerted by the magnetic field on the L-shaped metal machine part is approximately 11.44 N. This demonstrates the principle of how magnetic fields interact with current-carrying conductors, resulting in a force that depends not only on the current and magnetic field strength but also on the geometry of the conductor itself.