IDNLearn.com offers a user-friendly platform for finding and sharing answers. Join our knowledgeable community to find the answers you need for any topic or issue.
Sagot :
To determine the magnitude of the magnetic force on each wire and rank them accordingly, we need to use the relationship for the magnetic force acting on a current-carrying wire:
[tex]\[ F = I \cdot L \cdot B \cdot \sin(\theta) \][/tex]
Here, [tex]\( I \)[/tex] is the current, [tex]\( L \)[/tex] is the length of the wire in the magnetic field, [tex]\( B \)[/tex] is the magnetic field strength, and [tex]\( \theta \)[/tex] is the angle between the direction of the current and the magnetic field direction (which is due east in this problem). For simplicity in comparison, we can assume that [tex]\( L \)[/tex] and [tex]\( B \)[/tex] are constants. Therefore, the magnetic force [tex]\( F \)[/tex] will be proportional to [tex]\( I \cdot \sin(\theta) \)[/tex].
Let’s analyze each wire:
1. Wire A:
- Current, [tex]\( I_A = 15 \)[/tex] A
- Angle [tex]\( \theta_A = 45^{\circ} \)[/tex] south of east
- The sine of the angle, [tex]\( \sin(45^{\circ}) = \sqrt{2}/2 \approx 0.707 \)[/tex]
- Proportional force for wire A:
[tex]\[ F_A \propto 15 \cdot \sin(45^{\circ}) = 15 \cdot 0.707 = 10.61 \][/tex]
2. Wire B:
- Current, [tex]\( I_B = 18 \)[/tex] A
- Angle [tex]\( \theta_B = 180^{\circ} \)[/tex] (due west)
- The sine of the angle, [tex]\( \sin(180^{\circ}) = 0 \)[/tex]
- Proportional force for wire B:
[tex]\[ F_B \propto 18 \cdot \sin(180^{\circ}) = 18 \cdot 0 = 0 \][/tex]
3. Wire C:
- Current, [tex]\( I_C = 10 \)[/tex] A
- Angle [tex]\( \theta_C = 60^{\circ} \)[/tex] south of east
- The sine of the angle, [tex]\( \sin(60^{\circ}) = \sqrt{3}/2 \approx 0.866 \)[/tex]
- Proportional force for wire C:
[tex]\[ F_C \propto 10 \cdot \sin(60^{\circ}) = 10 \cdot 0.866 = 8.66 \][/tex]
Now, we compare the magnitudes of the proportional forces:
- [tex]\( F_B \approx 0 \)[/tex]
- [tex]\( F_C \approx 8.66 \)[/tex]
- [tex]\( F_A \approx 10.61 \)[/tex]
Ranking the wires in terms of the magnitude of the magnetic force, from smallest to largest:
[tex]\[ F_B < F_C < F_A \][/tex]
Thus, the correct order is:
[tex]\[ B < C < A \][/tex]
And the corresponding option is:
[tex]\[ \boxed{B < C < A} \][/tex]
[tex]\[ F = I \cdot L \cdot B \cdot \sin(\theta) \][/tex]
Here, [tex]\( I \)[/tex] is the current, [tex]\( L \)[/tex] is the length of the wire in the magnetic field, [tex]\( B \)[/tex] is the magnetic field strength, and [tex]\( \theta \)[/tex] is the angle between the direction of the current and the magnetic field direction (which is due east in this problem). For simplicity in comparison, we can assume that [tex]\( L \)[/tex] and [tex]\( B \)[/tex] are constants. Therefore, the magnetic force [tex]\( F \)[/tex] will be proportional to [tex]\( I \cdot \sin(\theta) \)[/tex].
Let’s analyze each wire:
1. Wire A:
- Current, [tex]\( I_A = 15 \)[/tex] A
- Angle [tex]\( \theta_A = 45^{\circ} \)[/tex] south of east
- The sine of the angle, [tex]\( \sin(45^{\circ}) = \sqrt{2}/2 \approx 0.707 \)[/tex]
- Proportional force for wire A:
[tex]\[ F_A \propto 15 \cdot \sin(45^{\circ}) = 15 \cdot 0.707 = 10.61 \][/tex]
2. Wire B:
- Current, [tex]\( I_B = 18 \)[/tex] A
- Angle [tex]\( \theta_B = 180^{\circ} \)[/tex] (due west)
- The sine of the angle, [tex]\( \sin(180^{\circ}) = 0 \)[/tex]
- Proportional force for wire B:
[tex]\[ F_B \propto 18 \cdot \sin(180^{\circ}) = 18 \cdot 0 = 0 \][/tex]
3. Wire C:
- Current, [tex]\( I_C = 10 \)[/tex] A
- Angle [tex]\( \theta_C = 60^{\circ} \)[/tex] south of east
- The sine of the angle, [tex]\( \sin(60^{\circ}) = \sqrt{3}/2 \approx 0.866 \)[/tex]
- Proportional force for wire C:
[tex]\[ F_C \propto 10 \cdot \sin(60^{\circ}) = 10 \cdot 0.866 = 8.66 \][/tex]
Now, we compare the magnitudes of the proportional forces:
- [tex]\( F_B \approx 0 \)[/tex]
- [tex]\( F_C \approx 8.66 \)[/tex]
- [tex]\( F_A \approx 10.61 \)[/tex]
Ranking the wires in terms of the magnitude of the magnetic force, from smallest to largest:
[tex]\[ F_B < F_C < F_A \][/tex]
Thus, the correct order is:
[tex]\[ B < C < A \][/tex]
And the corresponding option is:
[tex]\[ \boxed{B < C < A} \][/tex]
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. IDNLearn.com provides the best answers to your questions. Thank you for visiting, and come back soon for more helpful information.