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Chapter 20: Problem 3

A circular loop of radius \(12.0 \mathrm{~cm}\) is placed in a uniform magnetic field. (a) If the field is directed perpendicular to the plane of the loop and the magnetic flux through the loop is \(8.00 \times 10^{-3} \mathrm{~T} \cdot \mathrm{m}^{2}\), what is the strength of the magnetic field? (b) If the magnetic field is directed parallel to the plane of the loop, what is the magnetic flux through the loop?

Short Answer

Expert verified
a) Roughly \(0.178 \, T\). \n b) The magnetic flux is \(0 \, T.m^{2}\).

Step by step solution

01

Calculation of Magnetic Field when Perpendicular

Using the formula for magnetic flux: \[\Phi = BA\cos \theta\], in the first case where the field is perpendicular to the plane of the loop, the angle \(\theta\) between the area vector and magnetic field is \(0^{\circ}\) (since they're perpendicular), hence \(cos \theta = 1\). Re-arranging the equation for \(B\) we get \(B = \frac{\Phi}{A}\). The area \(A\) is the area of the loop, which is a circle with radius \(r = 12.0 \, cm = 0.12 \, m\). The area of a circle is given by \(\pi r^2\), so substituting the radius, \(A = \pi (0.12)^2\). Replacing \(\Phi = 8.00 \times 10^{-3} \, T.m^{2}\) and \(A\) in the equation will give the magnetic field strength \(B\).
02

Calculation of Magnetic Flux when Parallel

For the second case where the magnetic field is parallel to the plane of the loop, the angle \(\theta\) between the magnetic field and area of loop is \(90^{\circ}\), hence \(cos \theta = 0\). Now again applying the formula for magnetic flux \[\Phi = BA\cos \theta\], we see that the magnetic flux is \(0\) because \(cos(90^\circ)=0\), regardless of the values of \(B\) and \(A\). So no matter how strong the magnetic field is, if it's parallel to the plane, the magnetic flux through the loop will be zero.
03

Conclusion

The exercise was about how the orientation of a magnetic field with respect to an area, in this case a circle, affects the magnetic flux passing through the area. The stronger a perpendicular magnetic field is, the more magnetic flux it produces, and if the magnetic field is parallel to the area, it doesn't contribute to any magnetic flux at all.

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Key Concepts

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

Uniform Magnetic Field
Imagine a region where the magnetic field lines are equally spaced and parallel to one another; this depicts a uniform magnetic field. In such a field, the strength and direction of the field are consistent throughout. This is different from non-uniform fields where the field lines might converge, diverge, or be unevenly spaced, indicating changes in magnetic field intensity or direction.

Understanding uniform magnetic fields is crucial in physics because they provide a simplified model for studying magnetic effects. For instance, in the exercise involving a loop in a magnetic field, assuming uniformity allows us to use the magnetic flux formula without worrying about varying field strength across the loop's area. It's this simplicity that makes concepts like electromagnetic induction in a uniform field easier to understand and calculate.
Magnetic Field Strength
The magnetic field strength, typically symbolized by B, quantifies how strong a magnetic field is. It's measured in teslas (T) and is a vector quantity, which means it has both magnitude and direction.

The presence of a magnetic field strength is felt by charges moving through it, which experience a force, hence it's a critical factor in determining the magnetic flux through a given area. As illustrated in the exercise with a loop in a magnetic field, the formula \( B = \frac{\Phi}{A} \) connects magnetic field strength, B, with the magnetic flux, \( \Phi \), and the area, A. A higher magnetic field strength will result in greater magnetic flux if the orientation of the field with the area is optimal, that is, perpendicular.
Perpendicular and Parallel Magnetic Field Orientation
The effectiveness of a magnetic field in creating magnetic flux through an area is deeply influenced by its orientation relative to that area. When the field is perpendicular to the area, we see its maximum effect. This is because the component of the field that influences flux is at its greatest. Mathematically, this is seen in the term \( cos \theta \) in the magnetic flux formula, which equals 1 when the field is perpendicular.

In contrast, if the field is parallel to the area, there is no component of the field penetrating the area, therefore no flux is generated. This means \( cos \theta \) is 0 when the field is parallel, making the flux zero, regardless of the field's strength or the size of the area. This concept is particularly important and is vividly demonstrated in the exercise's solution, showing the stark difference in magnetic flux between perpendicular and parallel orientations.

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Most popular questions from this chapter

When the coil of a motor is rotating at maximum speed, the current in the windings is \(4.0 \mathrm{~A}\). When the motor is first turned on, the current in the windings is \(11 \mathrm{~A}\). If the motor is operated at \(120 \mathrm{~V}\), find \((a)\) the resistance of the windings and (b) the back emf in the coil at maximum speed.

The current in a coil changes from \(3.5 \mathrm{~A}\) to \(2.0 \mathrm{~A}\) in \(0.50 \mathrm{~s}\). If the average emf induced in the coil is \(12 \mathrm{mV}\), what is the self-inductance of the coil?

Considerable scientific work is currently under way to determine whether weak oscillating magnetic fields such as those found near outdoor electric power lines can affect human health. One study indicated that a magnetic field of magnitude \(1.0 \times 10^{-3} \mathrm{~T}\), oscillating at \(60 \mathrm{~Hz}\), might stimulate red blood cells to become cancerous. If the diameter of a red blood cell is \(8.0 \mu \mathrm{m}\), determine the maximum emf that can be generated around the perimeter of the cell.

An aluminum ring of radius \(5.00 \mathrm{~cm}\) and resistance \(3.00 \times 10^{-4} \Omega\) is placed around the top of a long air-core solenoid with 1000 turns per meter and a smaller radius of \(3.00 \mathrm{~cm}\), as in Figure \(\mathrm{P} 20.64\). If the current in the sole noid is increasing at a constant rate of \(270 \mathrm{~A} / \mathrm{s}\), what is the induced current in the ring? Assume the magnetic field produced by the solenoid over the area at the end of the solenoid is one-half as strong as the field at the center of the solenoid. Assume also the solenoid produces a negligible field outside its cross-sectional area.

The plane of a rectangular coil, \(5.0 \mathrm{~cm}\) by \(8.0 \mathrm{~cm}\), is perpendicular to the direction of a magnetic field \(\vec{B}\). If the coil has 75 turns and a total resistance of \(8.0 \Omega\), at what rate must the magnitude of \(\overrightarrow{\mathbf{B}}\) change to induce a current of \(0.10 \mathrm{~A}\) in the windings of the coil?
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