Question

In: Physics

In the figure below, two semicircular arcs have radii R2 = 7.90 cm and R1 = 2.75 cm, carry current i = 0.281 A, and share the same center of curvature C.

In the figure below, two semicircular arcs have radii R2 = 7.90 cm and R1 = 2.75 cm, carry current i = 0.281 A, and share the same center of curvature C.

(a) What is the magnitude of the net magnetic field at C?
______T

(b) What is the direction of the net magnetic field at C?
out of page or into page

Solutions

Expert Solution

Concepts and reason

The concept used to solve this problem is the magnetic field at the center of curvature of the semi-circular arcs. Initially, the magnetic field at point C due to inner radius can be calculated by using the relation of the magnetic field for a semi-circular loop. Later, the magnetic field at point C due to outer radius can be calculated by using the relation for the magnetic field for a semi-circular loop. Finally, the net magnetic field can be calculated by adding both magnetic fields.

Fundamentals

The magnetic field at the center of an arc curves through the angle \(\theta\). The Biot-Savart law describes the magnetic field created by a current-carrying conductor and allows calculating the strength of the field at various points. According to the Biot-Savart law, the magnetic field is represented as follows:

\(d B=\frac{\mu_{o} I d l}{4 \pi R^{2}}\)

Here, \(d B\) is the magnetic field of the small element, \(\mu_{0}\) is the permeability in free space, \(d l\) is the infinitesimal length of the conductor, and \(R\) is the radius. The arc length of the semi-circular curve is as follows:

\(s=R \theta\)

Here, \(\boldsymbol{\theta}\) is the angle. The expression for the net magnetic field is as follows:

\(B=B_{1}+B_{2}\)

Here, \(B\) is the net magnetic field.

(a) According to the Biot-Savart law, the magnetic field is represented as follows:

\(d B=\frac{\mu_{o} I d l}{4 \pi R^{2}}\)

Integrate the above relation to get the net field as follows:

\(B=\frac{\mu_{o} I}{4 \pi R^{2}} \int d l\)

$$ =\frac{\mu_{o} I s}{4 \pi R^{2}} $$

Here, \(s\) is the arc length. The expression for the magnetic field due to the inner radius is as follows:

\(B_{1}=\frac{\mu_{o} i s_{1}}{4 \pi R_{1}^{2}}\)

Here, \(B_{1}\) is the magnetic field due to inner radius, \(\mu_{o}\) is the permeability of free space, \(i\) is the current, \(s_{1}\) is the arc length

of the inner curve and \(R_{1}\) is the inner radius.

Replace \(R_{1} \theta\) for \(s_{1}\) and \(\theta\) for \(\pi\) in the above relation.

\(B_{1}=\frac{\mu_{o} i R_{1} \pi}{4 \pi R_{1}^{2}}\)

$$ =\frac{\mu_{o} i}{4 R_{1}} $$

Substitute \(0.281 \mathrm{~A}\) for \(i, 4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}\) for \(\mu_{o},\) and \(2.75 \mathrm{~cm}\) for \(R_{1}\)

$$ \begin{aligned} B_{1}=& \frac{\left(4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}\right)(0.281 \mathrm{~A})}{4\left[(2.75 \mathrm{~cm})\left(\frac{1 \times 10^{-2} \mathrm{~m}}{1 \mathrm{~m}}\right)\right]} \\ &=3.21 \times 10^{-6} \mathrm{~T} \end{aligned} $$

Explanation | Common mistakes | Hint for next step

The magnetic field due to the inner radius is determined by using the application of Biot-Savart law. The direction of the current flow is in positive direction.

The expression for the magnetic field due to the outer radius is as follows:

\(B_{2}=\frac{\mu_{o}(-i) s_{2}}{4 \pi R_{2}^{2}}\)

Here, \(B_{2}\) is the magnetic field due to outer radius, \(s_{2}\) is the arc length of the outer curve, and \(R_{2}\) is the outer radius.

Replace \(R_{2} \theta\) for \(s_{2}\) and \(\theta\) for \(\pi\) in the above relation.

\(B_{2}=\left(\frac{\mu_{o}}{4 \pi}\right)\left(\frac{R_{2}(-i) \pi}{R_{2}^{2}}\right)\)

$$ =-\frac{\mu_{o} i}{4 R_{2}} $$

Substitute \(0.281 \mathrm{~A}\) for \(i, 4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}\) for \(\mu_{o},\) and \(7.90 \mathrm{~cm}\) for \(R_{2}\)

$$ \begin{aligned} B_{2} &=-\frac{\left(4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}\right)(0.281 \mathrm{~A})}{4\left[(7.90 \mathrm{~cm})\left(\frac{1 \times 10^{-2} \mathrm{~m}}{1 \mathrm{~m}}\right)\right]} \\ &=-1.12 \times 10^{-6} \mathrm{~T} \end{aligned} $$

Explanation | Common mistakes | Hint for next step

The magnetic field due to the outer radius is determined using the concept of Biot-Savart law since the negative sign indicates that the direction of current is counter-clockwise. So that the magnetic field at the center of the semicircular arc is out of the page.

The expression for the net magnetic field is as follows:

$$ B=B_{1}+B_{2} $$

Substitute \(\left(-1.12 \times 10^{6} \mathrm{~T}\right)\) for \(B_{2}\) and \(3.21 \times 10^{-6} \mathrm{~T}\) for \(B_{1}\)

$$ \begin{aligned} B &=\left[\left(3.21 \times 10^{-6} \mathrm{~T}\right)-\left(1.12 \times 10^{-6} \mathrm{~T}\right)\right] \\ &=2.09 \times 10^{-6} \mathrm{~T} \\ & \approx 2.09 \mu \mathrm{T} \end{aligned} $$

Part a

Thus, the magnitude of the net magnetic field at the center of curvature of two semicircular arcs is \(2.09 \mu \mathrm{T}\).

Explanation | Common mistakes | Hint for next step

The required net magnetic field is calculated by using the value of the magnetic field due to inner and outer radius. A pure electromagnetic field cannot exist in the absence of curvature of circle. Because this curvature prevents the magnetic lines that crosses in the perfect field.

(b)

From part (a) calculations, the magnitude of the net magnetic field at the center of curvature is \(2.09 \mu \mathrm{T}\). Since the positive sign indicates that the current is clockwise, the direction of the magnetic field at the center of the semicircular arcs acts into the page.

Part b Thus, the direction of the net magnetic field at the center of curvature C acts into the page.

Explanation | Common mistakes

The direction of the magnetic field is determined using the right hand rule for the straight wire. But here the conductor is considered as the semicircular arcs. The magnetic field at the center of the curvature \(C\) acts into the page.


Part a

Thus, the magnitude of the net magnetic field at the center of curvature of two semicircular arcs is \(2.09 \mu T\).

Part \(b\)

Thus, the direction of the net magnetic field at the center of curvature C acts into the page.

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