A circular flexible loop of wire of radius r carrying a current I is placed in a uniform magnetic field B . If B is doubled, then tension in the loop

Options

A. remain unchanged
B. is doubled
C. is halved
D. becomes 4 times

Answer

To solve the problem of finding the tension in a circular flexible loop of wire carrying a current \( I \) when the uniform magnetic field \( B \) is doubled, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
We have a circular loop of wire with radius \( r \) carrying a current \( I \). The loop is placed in a uniform magnetic field \( B \). The loop is flexible, meaning it can slightly expand or contract.

2. **Force on the Loop**:
When a current-carrying wire is placed in a magnetic field, it experiences a magnetic force. The force \( F \) on a small segment of the wire can be expressed as:
\[ F = I \cdot L \cdot B \]
where \( L \) is the length of the wire segment in the magnetic field.

3. **Considering a Small Element of the Loop**:
For a small element of the loop, we can denote its length as \( dl \). The angle subtended at the center of the loop by this small segment is \( d\theta \). The length of this segment can be expressed as:
\[ dl = r \cdot d\theta \]

4. **Magnetic Force on the Element**:
The magnetic force acting on this small segment is then:
\[ dF = I \cdot (r \cdot d\theta) \cdot B \]

5. **Components of Tension**:
The tension \( T \) in the wire acts along the wire. For the small segment, the vertical components of the tension will balance the magnetic force. The vertical component of tension can be expressed as:
\[ 2T \sin\left(\frac{d\theta}{2}\right) \]
For small angles, \( \sin\left(\frac{d\theta}{2}\right) \approx \frac{d\theta}{2} \).

6. **Balancing Forces**:
Setting the upward force (from tension) equal to the downward force (from magnetic force):
\[ 2T \cdot \frac{d\theta}{2} = I \cdot (r \cdot d\theta) \cdot B \]
Simplifying gives:
\[ T \cdot d\theta = I \cdot r \cdot d\theta \cdot B \]
Dividing both sides by \( d\theta \) (assuming \( d\theta \neq 0 \)):
\[ T = I \cdot r \cdot B \]

7. **Effect of Doubling the Magnetic Field**:
If the magnetic field \( B \) is doubled, the new tension \( T' \) becomes:
\[ T' = I \cdot r \cdot (2B) = 2 \cdot (I \cdot r \cdot B) = 2T \]
Therefore, the tension in the loop doubles when the magnetic field is doubled.

### Final Answer:
The tension in the loop when the magnetic field \( B \) is doubled becomes \( 2T \).