`M/(Vr)` has the dimensions of (M = Magnetic moment, V = velocity , r = radius )

Options

A. Pole strength
B. Electric charge
C. Electric potential
D. Force

Answer

To solve the problem of determining the dimensions of the expression \( \frac{M}{Vr} \) where \( M \) is the magnetic moment, \( V \) is the velocity, and \( r \) is the radius, we will follow these steps:

### Step 1: Determine the dimensions of magnetic moment \( M \)
The magnetic moment \( M \) can be expressed in terms of current and area. The formula for magnetic moment is:
\[ M = I \cdot A \]
where \( I \) is the current and \( A \) is the area.

The dimensions of current \( I \) are:
\[ [I] = A \quad (\text{Ampere}) \]

The dimensions of area \( A \) (which is length squared) are:
\[ [A] = L^2 \]

Thus, the dimensions of magnetic moment \( M \) are:
\[ [M] = [I] \cdot [A] = A \cdot L^2 \]

### Step 2: Determine the dimensions of velocity \( V \)
Velocity is defined as distance divided by time:
\[ V = \frac{d}{t} \]
where \( d \) is distance (length) and \( t \) is time.

The dimensions of velocity \( V \) are:
\[ [V] = \frac{L}{T} \quad (L = \text{length}, T = \text{time}) \]

### Step 3: Determine the dimensions of radius \( r \)
The radius \( r \) is simply a measure of length, so its dimensions are:
\[ [r] = L \]

### Step 4: Combine the dimensions in the expression \( \frac{M}{Vr} \)
Now we substitute the dimensions we found into the expression \( \frac{M}{Vr} \):
\[ \frac{M}{Vr} = \frac{A \cdot L^2}{\left(\frac{L}{T}\right) \cdot L} \]

### Step 5: Simplify the expression
First, simplify the denominator:
\[ Vr = \left(\frac{L}{T}\right) \cdot L = \frac{L^2}{T} \]
Now substitute this back into the expression:
\[ \frac{M}{Vr} = \frac{A \cdot L^2}{\frac{L^2}{T}} = A \cdot L^2 \cdot \frac{T}{L^2} \]

The \( L^2 \) terms cancel out:
\[ \frac{M}{Vr} = A \cdot T \]

### Step 6: Identify the physical quantity
The dimensions \( A \cdot T \) correspond to electric charge \( Q \), since:
\[ Q = I \cdot T \]
where \( I \) has dimensions \( A \) (Ampere) and \( T \) is time.

### Conclusion
Thus, the dimensions of \( \frac{M}{Vr} \) correspond to electric charge.

### Final Answer
The expression \( \frac{M}{Vr} \) has the dimensions of electric charge.

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