A wire loop that encloses an area of `20 cm^2` has a resistance of `10 Omega` The loop is placed in a magnetic field of 2.4 T with its plane perpendicular to the field . The loop is suddenly removed from the field. How much charge flows past a given point in the wire ?

Options

A. `12xx10^(-4)C`
B. `3.6xx10^(-1)C`
C. `4.8xx10^(-4)C`
D. `2.4xx10^(-3)C`

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Convert the area from cm² to m²
The area \( A \) of the wire loop is given as \( 20 \, \text{cm}^2 \). We need to convert this to square meters.

\[ A = 20 \, \text{cm}^2 = 20 \times 10^{-4} \, \text{m}^2 = 2 \times 10^{-3} \, \text{m}^2 \]

### Step 2: Identify the magnetic field strength
The magnetic field strength \( B \) is given as \( 2.4 \, \text{T} \).

### Step 3: Calculate the initial magnetic flux
The magnetic flux \( \Phi_1 \) through the loop when it is in the magnetic field is calculated using the formula:

\[ \Phi_1 = B \cdot A \]

Substituting the values:

\[ \Phi_1 = 2.4 \, \text{T} \times 2 \times 10^{-3} \, \text{m}^2 = 4.8 \times 10^{-3} \, \text{Wb} \]

### Step 4: Calculate the final magnetic flux
When the loop is removed from the magnetic field, the magnetic flux \( \Phi_2 \) becomes zero:

\[ \Phi_2 = 0 \, \text{Wb} \]

### Step 5: Calculate the change in magnetic flux
The change in magnetic flux \( \Delta \Phi \) is given by:

\[ \Delta \Phi = \Phi_2 - \Phi_1 = 0 - 4.8 \times 10^{-3} \, \text{Wb} = -4.8 \times 10^{-3} \, \text{Wb} \]

### Step 6: Use Faraday's Law of Induction
According to Faraday's Law, the induced electromotive force (emf) \( E \) in the loop is given by:

\[ E = -\frac{\Delta \Phi}{\Delta t} \]

### Step 7: Relate induced emf to current
Using Ohm's Law, the induced emf can also be expressed as:

\[ E = I \cdot R \]

where \( I \) is the current and \( R \) is the resistance of the loop.

### Step 8: Express current in terms of charge
The current \( I \) can be expressed in terms of charge \( \Delta Q \):

\[ I = \frac{\Delta Q}{\Delta t} \]

### Step 9: Substitute and solve for charge
Substituting the expression for current into the equation for emf gives:

\[ -\frac{\Delta \Phi}{\Delta t} = \frac{\Delta Q}{\Delta t} \cdot R \]

This simplifies to:

\[ \Delta Q = -\frac{\Delta \Phi}{R} \]

Substituting the values we have:

\[ \Delta Q = -\frac{-4.8 \times 10^{-3} \, \text{Wb}}{10 \, \Omega} = \frac{4.8 \times 10^{-3}}{10} = 4.8 \times 10^{-4} \, \text{C} \]

### Final Answer
The charge that flows past a given point in the wire is:

\[ \Delta Q = 4.8 \times 10^{-4} \, \text{C} \]

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