A square loop of area `23cm^(2)` has a resistance of `10 Omega`.The loop is placed in uniform magnetic field of magnitude `40.0T`.The plane of loop is perpendicular to the magnetic field.The work done in pulling the loop out of the magnetic field slowly and uniformly in `1.0` sec,will be

Options

A. `5times10^(-3)J`
B. `1.0times10^(-4)J`
C. `2.5times10^(-3)J`
D. `1.0times10^(-3)J`

Answer

To solve the problem, we need to calculate the work done in pulling a square loop out of a magnetic field. Here’s a step-by-step breakdown of the solution:

### Step 1: Convert Area to SI Units
The area of the loop is given as \(23 \, \text{cm}^2\). We need to convert this to square meters.

\[ \text{Area} = 23 \, \text{cm}^2 = 23 \times 10^{-4} \, \text{m}^2 = 0.0023 \, \text{m}^2 \]

**Hint:** Remember to convert square centimeters to square meters by multiplying by \(10^{-4}\).

### Step 2: Calculate the Side Length of the Square Loop
Since the area \(A\) of a square loop is given by \(A = L^2\), we can find the side length \(L\).

\[ L = \sqrt{A} = \sqrt{0.0023} \approx 0.04796 \, \text{m} \]

**Hint:** To find the side length of a square from its area, take the square root of the area.

### Step 3: Calculate the Induced EMF
The induced electromotive force (EMF) in the loop when it is pulled out of the magnetic field can be calculated using Faraday's law of electromagnetic induction. The formula for EMF (\(\mathcal{E}\)) is:

\[ \mathcal{E} = B \cdot L \cdot v \]

Where:
- \(B = 40 \, \text{T}\) (magnetic field strength)
- \(L\) is the length of one side of the loop
- \(v\) is the velocity at which the loop is pulled out.

Given that the loop is pulled out uniformly in \(1.0 \, \text{s}\), we can assume a constant velocity:

\[ v = \frac{L}{t} = \frac{0.04796}{1.0} \approx 0.04796 \, \text{m/s} \]

Now substituting the values into the EMF formula:

\[ \mathcal{E} = 40 \cdot 0.04796 \cdot 0.04796 \approx 0.0916 \, \text{V} \]

**Hint:** The induced EMF can be calculated using the product of the magnetic field, the length of the loop, and the velocity.

### Step 4: Calculate the Current in the Loop
Using Ohm's law, the current \(I\) flowing through the loop can be calculated as:

\[ I = \frac{\mathcal{E}}{R} \]

Where \(R = 10 \, \Omega\):

\[ I = \frac{0.0916}{10} \approx 0.00916 \, \text{A} \]

**Hint:** Use Ohm's law to relate EMF, current, and resistance.

### Step 5: Calculate the Force on the Loop
The force \(F\) acting on the loop due to the magnetic field can be calculated using the formula:

\[ F = BIL \]

Substituting the values:

\[ F = 40 \cdot 0.00916 \cdot 0.04796 \approx 0.0175 \, \text{N} \]

**Hint:** The force on a current-carrying conductor in a magnetic field can be calculated using the product of the magnetic field, current, and length.

### Step 6: Calculate the Work Done
The work done \(W\) in pulling the loop out of the magnetic field is given by:

\[ W = F \cdot d \]

Where \(d\) is the distance moved, which is equal to the length of the side of the loop \(L\):

\[ W = F \cdot L = 0.0175 \cdot 0.04796 \approx 0.000839 \, \text{J} \]

**Hint:** Work done can be calculated as the product of force and the distance moved in the direction of the force.

### Final Result
The work done in pulling the loop out of the magnetic field slowly and uniformly in \(1.0 \, \text{s}\) is approximately:

\[ W \approx 0.000839 \, \text{J} \approx 8.39 \times 10^{-4} \, \text{J} \]