A uniform magnetic field of 1.5T exists in a cylinderical region of radius 10.0 cm, its direction being parallel to the axis along east to west. A current carrying wire in north-south direction passes through this region. The wire intersects the axis and experience a force of 1.2N downwards. If the wire is turned from north south to northest-southwest direction. then magnitude and direction of force is

Options

A. 1.2 N, upward
B. `1.2sqrt2`, downward
C. `1.2N`, downward
D. `1.2/(sqrt2)N`, downward

Explanation

To solve the problem, we need to analyze the situation step by step.

### Step 1: Understand the Initial Setup
We have a uniform magnetic field of 1.5 T directed from east to west. A current-carrying wire is oriented in the north-south direction and experiences a downward force of 1.2 N.

**Hint:** Visualize the magnetic field and the wire's orientation to understand how they interact.

### Step 2: Determine the Force on the Wire
The force on a current-carrying wire in a magnetic field is given by the formula:
\[ F = I \cdot B \cdot L \cdot \sin(\theta) \]
where:
- \( F \) is the force,
- \( I \) is the current,
- \( B \) is the magnetic field strength,
- \( L \) is the length of the wire in the magnetic field,
- \( \theta \) is the angle between the magnetic field and the current direction.

Initially, since the wire is in the north-south direction and the magnetic field is east-west, the angle \( \theta \) is 90 degrees. Thus, \( \sin(90^\circ) = 1 \).

### Step 3: Calculate the Current
From the initial force:
\[ F = I \cdot B \cdot L \]
Given \( F = 1.2 \, N \) and \( B = 1.5 \, T \), we can express the current:
\[ 1.2 = I \cdot 1.5 \cdot L \]
This implies:
\[ I \cdot L = \frac{1.2}{1.5} = 0.8 \]

### Step 4: Change the Wire Orientation
Now, the wire is turned from the north-south direction to the northeast-southwest direction. This means the angle \( \theta \) between the magnetic field and the current direction is now 45 degrees.

### Step 5: Calculate the New Force
Using the same formula for the force with the new angle:
\[ F' = I \cdot B \cdot L \cdot \sin(45^\circ) \]
Since \( \sin(45^\circ) = \frac{\sqrt{2}}{2} \), we have:
\[ F' = I \cdot B \cdot L \cdot \frac{\sqrt{2}}{2} \]

### Step 6: Substitute the Current Value
We already found \( I \cdot L = 0.8 \):
\[ F' = 0.8 \cdot 1.5 \cdot \frac{\sqrt{2}}{2} \]
Calculating this gives:
\[ F' = 0.8 \cdot 1.5 \cdot 0.7071 \approx 0.848 \, N \]

### Step 7: Determine the Direction of the Force
To find the direction of the force, we can use the right-hand rule. Curl your fingers from the direction of the current (northeast-southwest) towards the direction of the magnetic field (east-west). The thumb will point in the direction of the force.

Since the wire is now oriented at 45 degrees, the force will still be directed downwards, but the magnitude will be less than the initial force.

### Final Result
The magnitude of the force when the wire is turned to the northeast-southwest direction is approximately **0.848 N**, and the direction of the force remains **downwards**.

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