A toroid of n turns, mean radius R and cross-sectional radius a carries current I. It is placed on a horizontal table taken as x-y plane. Its magnetic moment `vecM`

Options

A. is non-zero and points in the z-direction by symmetry
B. points along the axis of the toroid (m=m`phi`)
C. is zero, otherwise there would be a field falling as `(1)/(r^(3))` at large distances outside the toroid
D. is pointing radially outwards.

Explanation

To find the magnetic moment \( \vec{M} \) of a toroid with \( n \) turns, mean radius \( R \), cross-sectional radius \( a \), and carrying current \( I \), we can follow these steps:

### Step 1: Understand the Structure of the Toroid
A toroid is essentially a circular loop of wire that is bent into a donut shape. It consists of multiple turns (or loops) of wire, each carrying the same current \( I \).

### Step 2: Calculate the Area of Each Current Loop
The area \( A \) of each circular loop can be calculated using the formula for the area of a circle:
\[ A = \pi a^2 \]
where \( a \) is the cross-sectional radius of the toroid.

### Step 3: Calculate the Magnetic Moment of Each Loop
The magnetic moment \( dM \) of a single loop can be expressed as:
\[ dM = I \cdot A \]
Substituting the area from Step 2:
\[ dM = I \cdot (\pi a^2) = I \pi a^2 \]

### Step 4: Calculate the Total Magnetic Moment for All Turns
Since there are \( n \) turns in the toroid, the total magnetic moment \( M \) can be calculated by multiplying the magnetic moment of one loop by the number of turns:
\[ M = n \cdot dM = n \cdot (I \pi a^2) = n I \pi a^2 \]

### Step 5: Consider the Direction of the Magnetic Moment
The direction of the magnetic moment for each loop is perpendicular to the plane of the loop. However, in a toroid, the loops are arranged such that the magnetic moments of diametrically opposite loops point in opposite directions.

### Step 6: Analyze the Cancellation of Magnetic Moments
Due to the symmetry of the toroid, the magnetic moments of diametrically opposite loops cancel each other out. Therefore, when you sum up all the magnetic moments, they effectively cancel out, leading to a total magnetic moment of:
\[ M = 0 \]

### Final Result
Thus, the net magnetic moment \( \vec{M} \) of the toroid is:
\[ \vec{M} = 0 \]