An ideal solenoid with \(N\) total turns has a current \(I\) passing through the helical wires that make up the solenoid. Ampère’s law is used with a rectangular path \(abcd\) as shown above to calculate the magnitude of the magnetic field \(B\) within the solenoid. The horizontal distances of the path are length \(x\), and the vertical distances are length \(y\).
Which of the following equations results from the correct application of Ampère’s law in this situation?
Correct Answer: \( \boxed{\mathrm{E}} \)
Ampère’s law states that
\( \displaystyle \oint \vec{B}\cdot d\vec{\ell}=\mu_0I_{\mathrm{enc}} \)
For an ideal solenoid, the magnetic field is:
• Uniform and parallel to the axis inside the solenoid.
• Negligible outside the solenoid.
Along segment \(bc\), the magnetic field is parallel to the path, so
\( \displaystyle \int_{bc}\vec{B}\cdot d\vec{\ell}=Bx \)
Along segment \(ad\), the field is approximately zero because it lies outside the solenoid.
Along segments \(ab\) and \(cd\), the path is perpendicular to the magnetic field, so
\( \vec{B}\cdot d\vec{\ell}=0 \).
Therefore, the line integral reduces to
\( Bx=\mu_0NI \)
Thus, the correct application of Ampère’s law gives \( \boxed{B(x)=\mu_0NI} \).
Therefore, the correct answer is (E).
A long straight wire of circular cross-section with radius \(a\) carries a uniform areal current density \(j\). What is the value of the line integral of the magnetic field around the circumference of the wire?
Correct Answer: \( \boxed{\mathrm{E}} \)
Ampère’s law states that
\( \displaystyle \oint \vec{B}\cdot d\vec{\ell}=\mu_0I_{\mathrm{enc}} \)
The enclosed current is equal to the current density multiplied by the cross-sectional area enclosed by the integration path.
Since the Amperian loop is taken around the circumference of the wire,
\( I_{\mathrm{enc}}=jA=j(\pi a^2) \)
Therefore,
\( \oint \vec{B}\cdot d\vec{\ell} =\mu_0I_{\mathrm{enc}} =\mu_0j\pi a^2 \)
Thus, the value of the line integral of the magnetic field around the wire is \( \boxed{\mu_0j\pi a^2} \).
Therefore, the correct answer is (E).
The current in a wire is \(5\,\mathrm{A}\). What is the value of the closed integral \( \oint \vec{B}\cdot d\vec{\ell} \) of the magnetic field along a closed path around the wire?
Correct Answer: \( \boxed{\mathrm{D}} \)
Apply Ampère’s law:
\( \oint \vec{B}\cdot d\vec{\ell}=\mu_0 I_{\mathrm{enc}} \)
The enclosed current is
\( I_{\mathrm{enc}}=5\,\mathrm{A} \)
Using
\( \mu_0=4\pi\times10^{-7}\ \mathrm{T\cdot m/A} \),
we obtain
\( \oint \vec{B}\cdot d\vec{\ell}=(4\pi\times10^{-7})(5) \)
\( =20\pi\times10^{-7}\ \mathrm{T\cdot m} \)
Therefore, the correct answer is (D).
