Question

Obtain the formula for the magnetic field inside a current carrying solenoid with the help of Ampere's circuital law.
OR
Obtain the formula for moment of couple acting on a magnetic dipole in a magnetic field. Define magnetic dipole moment with the help of this.

Hint:

For the solenoid apply \( \oint \vec{B}\cdot d\vec{l} = \mu_0 I_{enc} \) on a rectangular loop with enclosed current \( nLI \) to get \( B=\mu_0 nI \). For the dipole, the two pole forces form a couple giving \( \tau = MB\sin\theta \) with \( M = m\times 2l \).

Answer 1

Option 1: Magnetic field inside a solenoid (Ampere's circuital law)

Step 1 (Ampere's law): Ampere's circuital law states that the line integral of the magnetic field around any closed loop equals \( \mu_0 \) times the total current enclosed by that loop:
\[ \oint \vec{B}\cdot d\vec{l} = \mu_0\, I_{\text{enclosed}} \]
Step 2 (Field of a long solenoid): A long solenoid has \( n \) turns per unit length carrying current \( I \). Inside a long solenoid the field \( B \) is uniform and directed along the axis; just outside it is negligibly small (nearly zero).
Step 3 (Choose an Amperian loop): Take a rectangular loop \( abcd \) with side \( ab = L \) lying inside the solenoid parallel to the axis, side \( cd \) of the same length lying outside, and the two short sides \( bc \) and \( da \) perpendicular to the axis. Evaluate the integral along the four sides:
along \( ab \): \( \vec{B}\parallel d\vec{l} \Rightarrow \int \vec{B}\cdot d\vec{l} = BL \);
along \( cd \): \( B \approx 0 \Rightarrow \) contribution \( = 0 \);
along \( bc \) and \( da \): \( \vec{B}\perp d\vec{l} \) (or \( B=0 \)) \( \Rightarrow \) contribution \( = 0 \).
Hence \( \oint \vec{B}\cdot d\vec{l} = BL \).
Step 4 (Current enclosed): The length \( L \) of the loop encloses \( nL \) turns, each carrying current \( I \), so the total enclosed current is
\[ I_{\text{enclosed}} = (nL)\,I \]
Step 5 (Combine): Substituting into Ampere's law,
\[ BL = \mu_0 (nL) I \] Cancelling \( L \),
\[\boxed{B = \mu_0 n I}\]
where \( n \) is the number of turns per unit length. The field inside a long solenoid is uniform and independent of its diameter.

Option 2: Torque (moment of couple) on a magnetic dipole

Step 1 (Set-up): Consider a magnetic dipole (a small bar magnet) of pole strength \( m \) and length \( 2l \) placed in a uniform magnetic field \( \vec{B} \), with its axis making an angle \( \theta \) with the field.
Step 2 (Forces on the poles): The north pole (\( +m \)) experiences a force \( mB \) along \( \vec{B} \), and the south pole (\( -m \)) experiences a force \( mB \) opposite to \( \vec{B} \). These two equal, opposite and parallel forces form a couple; their resultant force is zero, so the dipole does not translate but only rotates.
Step 3 (Moment of the couple): Torque = (magnitude of either force) \( \times \) (perpendicular distance between the two forces). The perpendicular distance between the lines of action of the forces is \( 2l\sin\theta \). Therefore
\[ \tau = mB \times 2l\sin\theta = (m\cdot 2l)\,B\sin\theta \]
Step 4 (Introduce the magnetic moment): Define the magnetic dipole moment as \( M = m\times 2l \). Then
\[ \tau = M B \sin\theta \qquad\text{or in vector form}\qquad \vec{\tau} = \vec{M}\times\vec{B} \]
\[\boxed{\tau = MB\sin\theta}\]
Step 5 (Definition of magnetic dipole moment): Put \( \theta = 90^\circ \) and \( B = 1 \) unit; then \( \tau = M \). Hence the magnetic dipole moment is numerically equal to the torque experienced by the dipole when it is placed perpendicular to a uniform magnetic field of unit strength. It is also the product of pole strength and magnetic length, \( M = m\times 2l \), directed from the south pole to the north pole.