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Chapter 11: Problem 12

(a) A 0.750-m-long section of cable carrying current to a car starter motor makes an angle of \(60^{\circ}\) with the Earth's \(5.50 \times 10^{-5} \mathrm{~T}\) field. What is the current when the wire experiences a force of \(7.00 \times 10^{-3} \mathrm{~N} ?\) (b) If you run the wire between the poles of a strong horseshoe magnet, subjecting \(5.00 \mathrm{~cm}\) of it to a 1.75-T field, what force is exerted on this segment of wire?

Short Answer

Expert verified
The current in case (a) is found using the formula and the values provided, and for case (b), the force is calculated using the magnetic field, the length of the segment of the wire, and the current found in (a).

Step by step solution

01

Understanding the Magnetic Force on a Current-Carrying Conductor

The magnetic force on a current-carrying conductor can be calculated using the formula \(F=BIL\sin(\theta)\), where \(F\) is the magnetic force, \(B\) is the magnetic field strength, \(I\) is the current, \(L\) is the length of wire in the magnetic field, and \(\theta\) is the angle between the wire and the direction of the magnetic field.
02

Solving for Current (a)

Given the magnetic force \(F=7.00 \times 10^{-3} N\), magnetic field strength \(B=5.50 \times 10^{-5} T\), length of the wire in the field \(L=0.750 m\), and the angle \(\theta=60^\circ\), we can rearrange the formula to solve for the current \(I\) as follows: \[I = \frac{F}{B L \sin(\theta)}\].
03

Calculating the Current (a)

Substituting the given values into the rearranged formula, you get \(I = \frac{7.00 \times 10^{-3} N}{(5.50 \times 10^{-5} T)(0.750 m)\sin(60^\circ)}\). We'll compute the sine of the angle and proceed with the arithmetic.
04

Final Current Calculation (a)

Computing the sine of \(60^\circ\) gives \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\). Substituting and solving for \(I\), the calculation becomes: \[I = \frac{7.00 \times 10^{-3}}{(5.50 \times 10^{-5})(0.750) \left( \frac{\sqrt{3}}{2} \right)}\].
05

Calculation for Force (b)

The same magnetic force formula \(F=BIL\sin(\theta)\) is used, with \(\theta = 90^\circ\) in this case, as the wire is perpendicular to the magnetic field lines. The sine of a \(90^\circ\) angle is 1, so the formula simplifies to \(F = BIL\) for this part.
06

Force on Wire in a Strong Magnetic Field (b)

Given the stronger magnetic field \(B=1.75 T\) and a segment of wire of length \(L=5.00 cm\), which is \(0.05 m\), we can use the current from part (a) to calculate the force \(F\) as follows: \[F = (1.75 T)(I)(0.05 m)\].
07

Calculating the Force on the Wire (b)

With the value of \(I\) computed in the previous steps, simply multiply by \(B\) and \(L\) to find the force \(F\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Magnetic Field
Let's imagine a magnetic field as an influence that a magnet exerts on its surroundings. This field can be pictured as lines of force flowing from the magnet's north pole to its south pole. These invisible lines guide the behavior of magnetic materials and charged particles moving within the field.

An important aspect to remember is that the strength of the magnetic field, denoted as 'B', is measured in teslas (T), and it can vary in intensity. In our everyday experience, the Earth itself generates a magnetic field, which is relatively weak compared to the one produced by a strong horseshoe magnet. The stronger the field or the closer the lines are to each other, the greater the magnetic influence on surrounding objects.
Current in Physics
Current is the rate at which electric charge flows through a conductor, such as a wire. It is a crucial concept in understanding magnetic force as it directly influences the magnitude of force experienced by a current-carrying conductor in a magnetic field. Measured in amperes (A), current, symbolized as 'I', represents how many coulombs of charge pass through the conductor per second. A higher current means more charge is moving through, and consequently, it will interact more intensely with magnetic fields.
Lorentz Force
When delving into the magnetic force on current-carrying conductors, we encounter the Lorentz force. It's the combined electric and magnetic force on a point charge due to electromagnetic fields. For a current-carrying wire, the Lorentz force is specifically the magnetic part of this larger force.

The formula for determining the magnetic force on a conductor is given by F = BILsin(θ). 'F' stands for the force, 'B' is the magnetic field strength, 'I' is the current flowing through the wire, 'L' is the length of wire that's within the magnetic field, and 'θ' is the angle between the wire and the magnetic field's direction. The direction of the force is always perpendicular to both the wire's direction and the magnetic field lines.
Right-hand Rule
Finding the direction of the magnetic force can be simplified using the 'right-hand rule.' To apply this rule, you would point your thumb in the direction of the current ('I'), and your fingers in the direction of the magnetic field ('B'). Then, your palm points in the direction of the force on a positively charged particle, or extensionally, the direction of the force on the current-carrying wire.

However, it's crucial to remember that the actual force direction depends on whether the charges are positive or negative. This tool helps visualize the orientation of force without complex calculations, which is immensely helpful in both theoretical studies and practical applications in physics and electrical engineering.

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Most popular questions from this chapter

(a) A cosmic ray proton moving toward the Earth at \(5.00 \times 10^{7} \mathrm{~m} / \mathrm{s}\) experiences a magnetic force of \(1.70 \times 10^{-16} \mathrm{~N}\). What is the strength of the magnetic field if there is a \(45^{\circ}\) angle between it and the proton's velocity? (b) Is the value obtained in part (a) consistent with the known strength of the Earth's magnetic field on its surface? Discuss.

Explain why the magnetic field would not be unique (that is, not have a single value) at a point in space where magnetic field lines might cross. (Consider the direction of the field at such a point.)

Integrated Concepts (a) A 0.140-kg baseball, pitched at \(40.0 \mathrm{~m} / \mathrm{s}\) horizontally and perpendicular to the Earth's horizontal \(5.00 \times 10^{-5} \mathrm{~T}\) field, has a \(100-n C\) charge on it. What distance is it deflected from its path by the magnetic force, after traveling \(30.0 \mathrm{~m}\) horizontally? (b) Would you suggest this as a secret technique for a pitcher to throw curve balls?

Why would a magnetohydrodynamic drive work better in ocean water than in fresh water? Also, why would superconducting magnets be desirable?

(a) A 200-turn circular loop of radius \(50.0 \mathrm{~cm}\) is vertical, with its axis on an east-west line. A current of \(100 \mathrm{~A}\) circulates clockwise in the loop when viewed from the east. The Earth's field here is due north, parallel to the ground, with a strength of \(3.00 \times 10^{-5} \mathrm{~T}\). What are the direction and magnitude of the torque on the loop? (b) Does this device have any practical applications as a motor?
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