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A wire loop that encloses an area of 20 ...

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 ?

A

`12xx10^(-4)C`

B

`3.6xx10^(-1)C`

C

`4.8xx10^(-4)C`

D

`2.4xx10^(-3)C`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, we will use Faraday's law of electromagnetic induction and the relationship between induced electromotive force (emf), current, resistance, and charge. ### Step 1: Understand the problem We have a wire loop with: - Area \( A = 20 \, \text{cm}^2 = 20 \times 10^{-4} \, \text{m}^2 \) - Resistance \( R = 10 \, \Omega \) - Magnetic field \( B = 2.4 \, \text{T} \) The loop is removed from the magnetic field, which means the magnetic flux through the loop changes. ### Step 2: Calculate the initial magnetic flux The magnetic flux \( \Phi \) through the loop is given by: \[ \Phi = B \cdot A \] Substituting the values: \[ \Phi_{\text{initial}} = 2.4 \, \text{T} \times 20 \times 10^{-4} \, \text{m}^2 = 2.4 \times 20 \times 10^{-4} = 4.8 \times 10^{-4} \, \text{Wb} \] ### Step 3: Determine the final magnetic flux When the loop is removed from the magnetic field, the final magnetic flux \( \Phi_{\text{final}} \) is: \[ \Phi_{\text{final}} = 0 \, \text{Wb} \] ### Step 4: Calculate the change in magnetic flux The change in magnetic flux \( \Delta \Phi \) is: \[ \Delta \Phi = \Phi_{\text{final}} - \Phi_{\text{initial}} = 0 - 4.8 \times 10^{-4} = -4.8 \times 10^{-4} \, \text{Wb} \] ### Step 5: Calculate the induced emf According to Faraday's law, the induced emf \( \mathcal{E} \) is equal to the negative rate of change of magnetic flux: \[ \mathcal{E} = -\frac{\Delta \Phi}{\Delta t} \] For the purpose of finding the charge, we can consider the magnitude: \[ \mathcal{E} = \frac{|\Delta \Phi|}{\Delta t} \] ### Step 6: Calculate the current The current \( I \) induced in the loop can be calculated using Ohm's law: \[ I = \frac{\mathcal{E}}{R} \] ### Step 7: Relate current to charge The charge \( Q \) that flows through the wire can be expressed as: \[ Q = I \cdot \Delta t \] Substituting \( I \): \[ Q = \frac{\mathcal{E}}{R} \cdot \Delta t \] ### Step 8: Substitute for charge using change in flux Since we are interested in the total charge that flows, we can express it as: \[ Q = \frac{|\Delta \Phi|}{R} \] Substituting the values: \[ Q = \frac{4.8 \times 10^{-4} \, \text{Wb}}{10 \, \Omega} = 4.8 \times 10^{-5} \, \text{C} \] ### Final Answer The total charge that flows past a given point in the wire is: \[ Q = 4.8 \times 10^{-5} \, \text{C} \]
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