Chapter 27, Problem 21P

(a)

To determine

The magnitude of magnetic field at center of loop.

Answer to Problem 21P

The magnetic field at center of loop is $54.5\text{μT}$ .

Explanation of Solution

Given:

The radius of loop is $R=3.0\text{cm}$

The current is $I=2.6\text{A}$ .

Formula used:

The expression for magnetic field is given by,

  $B_{\text{x}}=\frac{{\mu }_0}{4\pi }\frac{2\pi R^{2}I}{{\left(x^2+R^2\right)}^{\frac{3}{2}}}$

Calculation:

The magnetic field at origin is calculated as,

  $\begin{array}{c}{B}_{\text{x}}=\frac{{\mu }_0}{4\pi }\frac{2\pi R^{2}I}{{\left(x^2+R^2\right)}^{\frac{3}{2}}}\\ =\left({10}^{-7}\text{N}/{\text{A}}^{\text{2}}\right)\frac{2\pi {\left(\left(3.0\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2\left(2.6\text{A}\right)}{{\left(x^2+\left(3.0\text{cm}\right){\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^2\right)}^{\frac{3}{2}}}\\ B_0=\left({10}^{-7}\text{N}/{\text{A}}^{\text{2}}\right)\frac{2\pi {\left(\left(3.0\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2\left(2.6\text{A}\right)}{{\left(0^2+\left(3.0\text{cm}\right){\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^2\right)}^{\frac{3}{2}}}\\ =5.45\times {10}^{-5}\text{T}\end{array}$

Further simplify the above,

  $\begin{array}{c}{B}_0=\left(\left(5.45\times {10}^{-5}\text{T}\right)\left(\frac{{10}^6\text{μT}}{1\text{T}}\right)\right)\\ =54.5\text{μT}\end{array}$

Conclusion:

Therefore, the magnetic field at center of loop is $54.5\text{μT}$ .

(b)

To determine

The magnitude of magnetic field at $1.0\text{cm}$ from the center.

Answer to Problem 21P

The magnetic field at center of loop is $46.5\text{μT}$ .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

  $\begin{array}{c}{B}_{\text{x}}=\frac{{\mu }_0}{4\pi }\frac{2\pi R^{2}I}{{\left(x^2+R^2\right)}^{\frac{3}{2}}}\\ =\left({10}^{-7}\text{N}/{\text{A}}^{\text{2}}\right)\frac{2\pi {\left(\left(3.0\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2\left(2.6\text{A}\right)}{{\left(x^2+\left(3.0\text{cm}\right){\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^2\right)}^{\frac{3}{2}}}\\ B_{0.01}=\left({10}^{-7}\text{N}/{\text{A}}^{\text{2}}\right)\frac{2\pi {\left(\left(3.0\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2\left(2.6\text{A}\right)}{{\left({\left(\left(1.0\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2+\left(3.0\text{cm}\right){\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^2\right)}^{\frac{3}{2}}}\\ =46.5\times {10}^{-6}\text{T}\end{array}$

Further simplify the above,

  $\begin{array}{c}{B}_0=\left(\left(46.5\times {10}^{-6}\text{T}\right)\left(\frac{{10}^6\text{μT}}{1\text{T}}\right)\right)\\ =46.5\text{μT}\end{array}$

Conclusion:

Therefore, the magnetic field at center of loop is $46.5\text{μT}$ .

(c)

To determine

The magnitude of magnetic field at $2.0\text{cm}$ from the center.

Answer to Problem 21P

The magnetic field at center of loop is $31.4\text{μT}$ .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

  $\begin{array}{c}{B}_{\text{x}}=\frac{{\mu }_0}{4\pi }\frac{2\pi R^{2}I}{{\left(x^2+R^2\right)}^{\frac{3}{2}}}\\ =\left({10}^{-7}\text{N}/{\text{A}}^{\text{2}}\right)\frac{2\pi {\left(\left(3.0\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2\left(2.6\text{A}\right)}{{\left(x^2+\left(3.0\text{cm}\right){\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^2\right)}^{\frac{3}{2}}}\\ B_{0.01}=\left({10}^{-7}\text{N}/{\text{A}}^{\text{2}}\right)\frac{2\pi {\left(\left(3.0\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2\left(2.6\text{A}\right)}{{\left({\left(\left(2.0\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2+\left(3.0\text{cm}\right){\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^2\right)}^{\frac{3}{2}}}\\ =31.4\times {10}^{-6}\text{T}\end{array}$

Further simplify the above,

  $\begin{array}{c}{B}_0=\left(\left(31.4\times {10}^{-6}\text{T}\right)\left(\frac{{10}^6\text{μT}}{1\text{T}}\right)\right)\\ =31.4\text{μT}\end{array}$

Conclusion:

Therefore, the magnetic field at center of loop is $31.4\text{μT}$ .

(d)

To determine

The magnitude of magnetic field at $35\text{cm}$ from the center.

Answer to Problem 21P

The magnetic field at center of loop is $33.9\text{nT}$ .

Explanation of Solution

Calculation:

The magnetic field is calculated as,

  $\begin{array}{c}{B}_{\text{x}}=\frac{{\mu }_0}{4\pi }\frac{2\pi R^{2}I}{{\left(x^2+R^2\right)}^{\frac{3}{2}}}\\ =\left({10}^{-7}\text{N}/{\text{A}}^{\text{2}}\right)\frac{2\pi {\left(\left(3.0\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2\left(2.6\text{A}\right)}{{\left(x^2+\left(3.0\text{cm}\right){\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^2\right)}^{\frac{3}{2}}}\\ B_{0.01}=\left({10}^{-7}\text{N}/{\text{A}}^{\text{2}}\right)\frac{2\pi {\left(\left(3.0\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2\left(2.6\text{A}\right)}{{\left({\left(\left(35\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2+\left(3.0\text{cm}\right){\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^2\right)}^{\frac{3}{2}}}\\ =33.9\times {10}^{-9}\text{T}\end{array}$

Further simplify the above,

  $\begin{array}{c}{B}_0=\left(\left(33.9\times {10}^{-9}\text{T}\right)\left(\frac{{10}^9\text{nT}}{1\text{T}}\right)\right)\\ =33.9\text{nT}\end{array}$

Conclusion:

Therefore, the magnetic field at center of loop is $33.9\text{nT}$ .