A beam of neutrons performs circular motion of radius, r = 1 m, under the influence of an inhomogeneous magnetic field with inhomogeneity extending over `Delta r = 0.01` m. The speed of the neutrons is 54 m/s. The mass and magnetic moment of the neutrons respectively are `1.67 xx 10^(–27) kg and 9.67 x× 10^(–27)` J/T. The average variation of the magnetic field over Dr is approximately.

Options

A. `0.5 T`
B. `1.0 T`
C. `5.0 T`
D. `10.0 T `

Explanation

To solve the problem, we need to calculate the average variation of the magnetic field over the given inhomogeneity (Δr). We will use the relationship between the magnetic force acting on the neutrons and the centripetal force required for circular motion.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Radius of circular motion, \( r = 1 \, \text{m} \)
- Inhomogeneity, \( \Delta r = 0.01 \, \text{m} \)
- Speed of neutrons, \( v = 54 \, \text{m/s} \)
- Mass of neutron, \( m_n = 1.67 \times 10^{-27} \, \text{kg} \)
- Magnetic moment of neutron, \( \mu = 9.67 \times 10^{-27} \, \text{J/T} \)

2. **Calculate the Centripetal Force:**
The centripetal force \( F_c \) required for circular motion is given by:
\[ F_c = \frac{m_n v^2}{r} \]
Substituting the values:
\[ F_c = \frac{(1.67 \times 10^{-27} \, \text{kg}) \times (54 \, \text{m/s})^2}{1 \, \text{m}} \]

3. **Calculate the Centripetal Force:**
First, calculate \( v^2 \):
\[ v^2 = 54^2 = 2916 \, \text{m}^2/\text{s}^2 \]
Now substitute this back into the centripetal force equation:
\[ F_c = \frac{(1.67 \times 10^{-27}) \times 2916}{1} \]
\[ F_c = 4.87 \times 10^{-24} \, \text{N} \]

4. **Relate the Centripetal Force to the Magnetic Force:**
The magnetic force \( F_m \) acting on the neutrons in a magnetic field \( B \) is given by:
\[ F_m = \mu B \]
For the average variation of the magnetic field over the inhomogeneity \( \Delta r \), we can express it as:
\[ F_m = \mu \Delta B \]
where \( \Delta B \) is the change in the magnetic field over the distance \( \Delta r \).

5. **Set the Forces Equal:**
Since the magnetic force provides the necessary centripetal force, we can set them equal:
\[ \mu \Delta B = F_c \]
Rearranging gives:
\[ \Delta B = \frac{F_c}{\mu} \]

6. **Calculate the Average Variation of the Magnetic Field:**
Substitute \( F_c \) and \( \mu \):
\[ \Delta B = \frac{4.87 \times 10^{-24}}{9.67 \times 10^{-27}} \]
\[ \Delta B \approx 5.03 \, \text{T} \]

### Final Answer:
The average variation of the magnetic field over \( \Delta r \) is approximately \( 5.03 \, \text{T} \).