An aluminum wire of length 1.5 m and mass of 0.070 kg is in a magnetic field of 0.40 T. What is the current in the wire such that the magnetic force balances the weight of the rod? 1.2 A 1.14 A 9.8 A 2.5 A
An aluminum wire of length 1.5 m and mass of 0.070 kg is in a magnetic field of 0.40 T. What is the current in the wire such that the magnetic force balances the weight of the rod? 1.2 A 1.14 A 9.8 A 2.5 A
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter29: Magnetic Fields
Section: Chapter Questions
Problem 29.12OQ
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![**Question:**
An aluminum wire of length 1.5 m and mass of 0.070 kg is in a magnetic field of 0.40 T. What is the current in the wire such that the magnetic force balances the weight of the rod?
**Options:**
- 1.2 A
- 1.14 A
- 9.8 A
- 2.5 A
**Explanation:**
This problem explores the concept of magnetic force on a current-carrying conductor in a magnetic field. The force can be calculated using the formula:
\[ F = B \times I \times L \]
where:
- \( F \) is the magnetic force,
- \( B \) is the magnetic field strength (0.40 T),
- \( I \) is the current in the wire,
- \( L \) is the length of the wire (1.5 m).
The gravitational force acting on the wire, which is the weight, is given by:
\[ F_{gravity} = m \times g \]
where:
- \( m \) is the mass of the wire (0.070 kg),
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).
To balance the weight of the rod with the magnetic force, we set \( F = F_{gravity} \).
Calculating \( F_{gravity} \):
\[ F_{gravity} = 0.070 \times 9.81 = 0.6867 \, \text{N} \]
Set the magnetic force equal to gravitational force:
\[ B \times I \times L = 0.6867 \]
\[ 0.40 \times I \times 1.5 = 0.6867 \]
Solve for \( I \):
\[ I = \frac{0.6867}{0.40 \times 1.5} \]
\[ I = \frac{0.6867}{0.6} \]
\[ I \approx 1.14 \, \text{A} \]
**Answer: 1.14 A**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2c067e04-819e-445d-9345-defba0ea4f93%2F6e40f718-ccf1-402a-89a6-a0f4542119ed%2F04p1mch_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question:**
An aluminum wire of length 1.5 m and mass of 0.070 kg is in a magnetic field of 0.40 T. What is the current in the wire such that the magnetic force balances the weight of the rod?
**Options:**
- 1.2 A
- 1.14 A
- 9.8 A
- 2.5 A
**Explanation:**
This problem explores the concept of magnetic force on a current-carrying conductor in a magnetic field. The force can be calculated using the formula:
\[ F = B \times I \times L \]
where:
- \( F \) is the magnetic force,
- \( B \) is the magnetic field strength (0.40 T),
- \( I \) is the current in the wire,
- \( L \) is the length of the wire (1.5 m).
The gravitational force acting on the wire, which is the weight, is given by:
\[ F_{gravity} = m \times g \]
where:
- \( m \) is the mass of the wire (0.070 kg),
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).
To balance the weight of the rod with the magnetic force, we set \( F = F_{gravity} \).
Calculating \( F_{gravity} \):
\[ F_{gravity} = 0.070 \times 9.81 = 0.6867 \, \text{N} \]
Set the magnetic force equal to gravitational force:
\[ B \times I \times L = 0.6867 \]
\[ 0.40 \times I \times 1.5 = 0.6867 \]
Solve for \( I \):
\[ I = \frac{0.6867}{0.40 \times 1.5} \]
\[ I = \frac{0.6867}{0.6} \]
\[ I \approx 1.14 \, \text{A} \]
**Answer: 1.14 A**
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