StudySmarter AI is coming soon!

- :00Days
- :00Hours
- :00Mins
- 00Seconds

A new era for learning is coming soonSign up for free

Suggested languages for you:

Americas

Europe

P 28

Expert-verifiedFound in: Page 836

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Question: Figure 29-56 a shows two wires, each carrying a current .Wire 1 consists of a circular arc of radius R and two radial lengths; it carries current ${{\mathit{i}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{}}{\mathbf{2}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{\text{A}}}$ in the direction indicated. Wire 2 is long and straight; it carries a current i_{2} that can be varied; and it is at distance **$\frac{\mathbf{R}}{\mathbf{2}}$

The angle subtended by the arc is $\theta =1.00\text{rad}$.

i) The current flowing through the wire is ${i}_{1}=2.00A$.

ii) The horizontal scale is set by ${i}_{2x}=1.00\text{A}$

ii) The radius of the circular arc is R.

iv) The distance between wire 2 and the center of the circular arc of wire 1 is $\frac{R}{2}$.

Use the concept of the magnetic force due to current in straight wires and current in circular arc.

**Formulae:**

${B}_{straight}=\frac{{\mu}_{0}i}{2\pi R}$

${B}_{arc}=\frac{{\mu}_{0}i\varnothing}{4\pi R}$

The angle subtended by the arc:

The magnetic field due to a current in long straight wire is

${B}_{straight}=\frac{{\mu}_{0}i}{2\pi R}$

Here, from the figure,

${B}_{straight}=\frac{{\mu}_{0}{i}_{2}}{2\pi \left(\frac{R}{2}\right)}$

According to the right hand, both give direction of magnetic field pointing out of the page.

The magnetic field due to the current in circular arc of the wire is

${B}_{arc}=\frac{{\mu}_{0}{i}_{1}\varnothing}{4\pi R}$

In the figure (a), according to the right hand, both wires give the direction of magnetic field pointing out of the page.

The net magnetic field is

$B={B}_{arc}-{B}_{straight}$

$B=\frac{{\mu}_{0}{i}_{1}\varnothing}{4\pi R}-\frac{{\mu}_{0}{i}_{2}}{2\pi \left(\frac{R}{2}\right)}$ …… (1)

From the figure (b), for ${i}_{2x}=0.5\text{A,}$ then $B=0\text{T}$

Hence, equation (1) becomes

$0=\frac{{\mu}_{0}{i}_{1}\varnothing}{4\pi R}-\frac{{\mu}_{0}{i}_{2}}{2\pi \left(\frac{R}{2}\right)}$

$\frac{{\mu}_{0}{i}_{1}\varnothing}{4\pi R}=\frac{{\mu}_{0}{i}_{2}}{2\pi \left(\frac{R}{2}\right)}$

$\frac{{\mu}_{0}{i}_{1}\varnothing}{4\pi R}=\frac{{\mu}_{0}{i}_{2}}{2\pi \left(\frac{R}{2}\right)}$

Solve further as:

role="math" localid="1663153369320" $\varnothing =4\left(\frac{{i}_{2}}{{i}_{1}}\right)$

$\varnothing =4\left(\frac{0.50A}{2.00A}\right)$

$\varnothing =1.00\text{rad}$

94% of StudySmarter users get better grades.

Sign up for free