Physics Problems & Solutions: Biot Savart's Law

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Electromagnetism - Biot Savart's Law

A current i in a circular loop of radius b produces a magnetic field. At a fixed points far from the loop, the strength of the magnetic field is proportional to which of the following combinations of i and b?

A. ib
B. ib²
C. i²b
D. i/b
E. i/b²

(GR8677 #65)

Solution:

Applying the concept of magnetic dipole moment:
μ = a vector quantity associated with the torque exerted by an external magnetic field on a current carrying coil

μ = IA = Iπr² = iπb²

Answer: B

Note: click HERE for complete calculation to show that magnetic field at a fixed distance x far away from a circular loop: $B= \frac{\mu_0 IR^2}{2x^3}$

Electromagnetism - Faraday’s law

A small circular wire loop of radius a is located at the center of a much larger circular wire loop radius b as shown above. The larger loop carries an alternating current I = I₀ cos ωt, where I₀ and ω are constants. The magnetic field generated by the current in the large loop induces in the small loop an emf that is approximately equal to which of the following? (Either use mks units and let μ₀ be the permeability of free space, or use Gaussian units and let μ₀ be 4π/c².)

A. $\left(\frac{\pi\mu_0 I_0}{2}\right)\frac{a^2}{b}\omega \cos \omega t$
B. $\left(\frac{\pi\mu_0 I_0}{2}\right)\frac{a^2}{b}\omega \sin \omega t$
C. $\left(\frac{\pi\mu_0 I_0}{2}\right)\frac{a}{b^2}\omega \sin \omega t$
D. $\left(\frac{\pi\mu_0 I_0}{2}\right)\frac{a}{b^2}\omega \cos \omega t$
E. $\left(\frac{\pi\mu_0 I_0}{2}\right)\frac{a}{b}\omega \sin \omega t$

(GR8677 #81)

Solution:

Faraday's Law: ɛ = − dΦ/dt
with Φ = NBA

Biot Savart Law: B ~ I

Given:
I = I₀ cos ωt → B ~ I₀ cos ωt
Area of smaller loop with radius a, A = πa²
N = 1

→ Φ = BA ≈ πa² I₀ cos ωt
→ ɛ ~ dΦ/dt ≈ πa² ω I₀ sin ωt

Only (B) fits the equation.

Answer: B

Complete Calculation:

Faraday's Law: ɛ = − dΦ/dt
with Φ = NBA

Magnetic field at the center of a current wire loop: $B= \frac{\mu_0 I}{2r}$ (Proof)
For the larger loop with radius b, carrying I = I₀ cos ωt → $B= \frac{\mu_0 I_0 \cos \omega t}{2b}$

Area of smaller loop with radius a → A = πa²

N = 1

$\rightarrow \varepsilon\approx\frac{d\Phi}{dt}= A \frac{dB}{dt}= (\pi a^2)\left( \frac{\mu_0 I_0}{2b}\right )\frac{d\cos \omega t}{dt}=\left(\frac{\pi\mu_0 I_0}{2}\right)\frac{a^2}{b}\omega \sin \omega t$

Electromagnetism - Biot Savart's Law

A segment of wire is bent into an arc of radius R and subtended angle θ, as shown in the figure. Point P is at the center of the circular segment. The wire carries current I. What is the magnitude of the magnetic field at P?

A. 0
B. ^μ₀I^θ/₍₂_π₎_²_R

C. ^μ₀I^θ/₄_πR
D. ^μ₀I^θ/₄_πR_²
E. ^μ₀I/₂_θ_R_²

(GR0177 #88)

Solution:

Biot Savart’s Law:

$\begin{aligned} d \vec{B}&=\frac{\mu_0}{4\pi} \frac{Id \vec{l}\times \hat{r}}{r^2} \\\\\hat{r}&=\vec{r}/r \\d \vec{B}&=\frac{\mu_0}{4\pi} \frac{Id \vec{l}\times \vec{r}}{r^3} \end{aligned}$

$\begin{aligned} &Id \vec{l}\perp \vec{R} \rightarrow Id \vec{l}\times\vec{R}=IR dl \\ &d \vec{B}=\frac{\mu_0}{4\pi} \frac{IR dl}{R^3} \\&dl=Rd\theta\rightarrow d \vec{B}=\frac{\mu_0}{4\pi} \frac{IR^2d\theta}{R^3} \\&B=\frac{\mu_0 I\theta}{4\pi R} \end{aligned}$

Answer: C

Biot Savart's Law - Magnetic Field for Circular Wire Loop

Show that the magnetic field at the center of the circular loop is $B= \frac{\mu_0 I}{2R}$ and at fixed distance x far away from the loop is $B= \frac{\mu_0 IR^2}{2x^3}$

Solution:

Biot Savart's Law: $d\vec{B}= \frac{\mu_0}{4\pi}\frac{Id\vec{l}\times \hat{r}}{r^2}$
$\hat r=\frac{\vec r}{r}$
and from the sketch: $Id\vec l\perp \vec r \rightarrow I d\vec l \times \vec r = I dl r \sin 90 = I dl r$

$\rightarrow d\vec{B}= \frac{\mu_0}{4\pi}\frac{Id\vec{l}\times \vec{r}}{r^3} = \frac{\mu_0}{4\pi}\frac{Idlr}{r^3}=\frac{\mu_0}{4\pi}\frac{Idl}{r^2}$

By symmetry, total dB_y = 0

$\rightarrow d \vec B = \int dB_x = \int d\vec B \sin \theta$

From the sketch: $\sin \theta = R/r$

$\rightarrow B(x)= \int \frac{\mu_0}{4\pi}\frac{Idl}{r^2} \frac{R}{r} =\frac{\mu_0 I R}{4\pi r^3}\oint dl = \frac{\mu_0 I R}{4\pi r^3} 2 \pi R=\frac{\mu_0 I R^2}{2 r^3}$

From the sketch: $r=\sqrt{R^2+x^2}$

$\rightarrow B(x)= \frac{\mu_0 I R^2}{2 (R^2+x^2)^{3/2}}$

At the center of the loop, x = 0 $\rightarrow B(x)= \frac{\mu_0 I}{2R}$
At fixed distance x far away from the loop, x ≫ R $\rightarrow B= \frac{\mu_0 IR^2}{2x^3}$