Solved Problem on Magnetic Field

A compass is placed inside an XY horizontal axis solenoid. The horizontal component of the earth's magnetic field is B_h and the solenoid has N turns of length ℓ. The solenoid axis is arranged perpendicular to the direction of the compass needle. Calculate the angle α described by the needle when the solenoid is traversed by a current of intensity i in the direction shown in the figure.

Problem data:

Solenoid length: ℓ;
Number of turns of the solenoid: N;
Solenoid current: i.

Problem diagram:

The current i produces in the solenoid a magnetic field given by the Right Hand Rule. By placing the fingers in the direction of the current, the thumb will indicate the direction of the Magnetic Field lines from Y (South Pole) to X (North Pole) (Figure 1).

Figure 1

Solution

The magnetic field of a solenoid is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {B_{S}=\mu_{0}\frac{N}{\ell}i} \tag{I} \end{gather} \]

Under the action of the magnetic field of the solenoid the needle moves through an angle α, the resultant magnetic field vector is given by (Figure 2-A)

\[ \begin{gather} \bbox[#99CCFF,10px] {\vec{B}={\vec{B}}_{S}+{\vec{B}}_{h}} \end{gather} \]

Figure 2

The vectors \( \vec{B} \), \( {\vec{B}}_{S} \) and \( {\vec{B}}_{h} \) form a right triangle (Figure 2-B), with hypotenuse equal to \( \vec{B} \) and legs equal to \( {\vec{B}}_{S} \) and \( {\vec{B}}_{h} \), the angle a is calculated by the tangent of α given by

\[ \begin{gather} \tan{\mathit\alpha}=\frac{B_{S}}{B_{h}} \tag{II} \end{gather} \]

substituting equation (I) into equation (II)

\[ \begin{gather} \tan{\mathit\alpha}=\mu_{0}\frac{N}{\ell}i\frac{1}{B_{h}} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {\mathit\alpha=\arctan\left(\mu_{0}\frac{N}{\ell}\frac{i}{B_{h}}\right)} \end{gather} \]

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