A long solenoid carries a steady current. An Amperian loop is chosen completely outside the solenoid. What is the magnetic field along this loop according to the ideal solenoid model?
Zero
In an ideal (infinitely long, tightly wound) solenoid, the magnetic field inside is uniform, \(B = \mu_0 n I\), directed along the axis.
For an Amperian loop lying entirely outside the solenoid, no current passes through the loop, so \(\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc} = 0\). The field lines that leave one end return over an effectively infinite region, so the external field is taken as zero.
The other options fail: the field is maximum and uniform only inside the solenoid; outside it is neither equal to the inside value nor a uniform non-zero field in the ideal model.
Hence, the answer is Zero.
The magnetic field inside a long solenoid is 4 × 10−3 T when it carries a current of 2 A. If the current is increased to 6 A, the magnetic field becomes:
An Amperian loop encloses three conductors carrying currents of 4 A, 3 A, and 1 A. Two currents are directed into the page and one is directed out of the page. Which additional information is required to determine the enclosed current?
Ampere's circuital law relates the line integral of the magnetic field around a closed path to:
The wire loop shown in the figure carries a steady current $I$. Each straight section of the loop has length $d$. A part of the loop lies in xy plane and the other part is tilted at $30^\circ$ with respect to the xz plane. The magnitude of the magnetic dipole moment of the loop (in appropriate units) is

An infinitely long cylinder of radius $R$ carries a frozen-in magnetization $\vec M = ke^{-s^2} \hat z$, where $k$ is a constant and $s$ is the distance from the axis of cylinder. The magnetic permeability of free space is $\mu_o$. There is no free current present anywhere. The magnetic flux density ($\vec B$) inside the cylinder is