Problem 36
Question
A straight, vertical wire carries a current of 1.20 \(\mathrm{A}\) downward in a region between the poles of a large superconducting electromagnet, where the magnetic field has magnitude \(B=\) 0.588 \(\mathrm{T}\) and is horizontal. What are the magnitude and direction of the magnetic force on a \(1.00-\mathrm{cm}\) section of the wire that is in this uniform magnetic field, if the magnetic field direction is (a) east; (b) south; (c) \(30.0^{\circ}\) south of west?
Step-by-Step Solution
Verified Answer
(a) 0.00706 N south, (b) 0.00706 N east, (c) 0.00612 N north-east.
1Step 1: Understanding the problem
We need to find the magnetic force on a vertical wire carrying a current of 1.20 A for a 1.00-cm section in different orientations of a uniform magnetic field of 0.588 T. The direction of the current is down the wire. We will use the magnetic force formula.
2Step 2: Using the magnetic force equation
The magnetic force experienced by a current-carrying wire in a magnetic field is given by \[ F = I \cdot L \cdot B \cdot \sin(\theta) \]where:- \( F \) is the magnetic force,- \( I \) is the current (1.20 A),- \( L \) is the length of the wire (1.00 cm converted to meters: 0.01 m),- \( B \) is the magnetic field strength (0.588 T),- \( \theta \) is the angle between the direction of the current and the magnetic field.
3Step 3: Calculating the force for case (a) - Magnetic field direction east
For part (a), the magnetic field is directed east. The current is downward (vertically), so the angle \( \theta \) between the current and magnetic field is 90 degrees. Using \( \sin(90^\circ) = 1 \), the force \( F \) is \[ F = 1.20 \cdot 0.01 \cdot 0.588 \cdot 1 = 0.007056 \, \text{N} \]The direction of the force is given by the right-hand rule, which points south.
4Step 4: Calculating the force for case (b) - Magnetic field direction south
In part (b), the magnetic field is directed south. The angle between the current and the magnetic field remains 90 degrees. Thus, \[ F = 1.20 \cdot 0.01 \cdot 0.588 \cdot 1 = 0.007056 \, \text{N} \]The right-hand rule indicates the magnetic force direction is east.
5Step 5: Calculating the force for case (c) - Magnetic field direction 30 degrees south of west
For part (c), the magnetic field makes an angle of 30 degrees south of west. The angle \( \theta \) between the current and the magnetic field is 90 + 30 = 120 degrees.Using \( \sin(120^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2} \),\[ F = 1.20 \cdot 0.01 \cdot 0.588 \cdot \frac{\sqrt{3}}{2} = 0.006116 \, \text{N} \]Using the right-hand rule, the direction of the force is north-east.
Key Concepts
Right-Hand RuleMagnetic Field DirectionCurrent-Carrying Wire in a Magnetic Field
Right-Hand Rule
The right-hand rule is a simple yet powerful tool used to determine the direction of the magnetic force on a current-carrying wire in a magnetic field. To apply this rule, extend your right hand so that your thumb, index, and middle fingers are perpendicular to each other.
- Your thumb represents the direction of the conventional current, which is the flow of positive charges.
- Your index finger indicates the direction of the magnetic field.
- Your middle finger then points in the direction of the force experienced by the wire.
Magnetic Field Direction
Understanding the direction of the magnetic field is crucial when calculating the magnetic force on a current-carrying wire. The magnetic field (\( B \)) is a vector field, meaning it has both magnitude and direction.It's represented by lines starting from the north pole of a magnet to the south pole. The direction of the magnetic field is tangent to the field lines at any point. In the exercise provided, the magnetic field is described in specific directions: east, south, or at an angle "30 degrees south of west."By knowing these orientations, you can apply the equation for magnetic force, \( F = I \cdot L \cdot B \cdot \sin(\theta) \), and correctly compute the force by considering the angle \( \theta \)between the wire and the field direction. For example, when the field is east and the wire's current direction is vertical, the angle is 90 degrees, giving an optimal interaction for calculating force.
Current-Carrying Wire in a Magnetic Field
A current-carrying wire in a magnetic field experiences a force due to the interaction between the current and the magnetic field. This force is calculated using the equation \( F = I \cdot L \cdot B \cdot \sin(\theta) \), which factors in the current \( I \), length of the wire \( L \), magnetic field \( B \), and the sine of the angle \( \theta \)between the current's and field's directions.In the given exercise, the wire is oriented vertically, carrying a current of 1.20 A. Depending on the magnetic field’s direction, the angle \( \theta \) changes which affects the force magnitude and direction.
- If the magnetic field is east, making \( \theta = 90^\circ \), the force direction is perpendicular to both the field and current, pointing south.
- With a field southward, the orientation again results in \( \theta = 90^\circ\), changing the force direction to east.
- When the field is 30 degrees south of west, \( \theta = 120^\circ\), creating a force towards the northeast.
Other exercises in this chapter
Problem 33
A straight \(2.00-\mathrm{m}, 150-\mathrm{g}\) wire carries a current in a region where the earth's magnetic field is horizontal with a magnitude of 0.55 gauss.
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An electromagnet produces a magnetic field of 0.550 T in a cylindrical region of radius 2.50 \(\mathrm{cm}\) between its poles. A straight wire carrying a curre
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A horizontal rod 0.200 \(\mathrm{m}\) long is mounted on a balance and carries a current. At the location of the rod a uniform horizontal magnetic field has mag
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\(\mathbf{A}\) thin 50.0 -cm-long metal bar with mass 750 g rests on, but is not attached to, two metallic supports in a uniform 0.450 - T magnetic field, as sh
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