The current in a wire varies with time according to the equation l=4+2t, where l is in ampere and t is in sec. the quantity of charge which has passed through a cross-section of the wire during the time t=2 sec to t=6 sec will be

Options

A. 40C
B. 48C
C. 38C
D. 43C

Explanation

To find the quantity of charge that has passed through a cross-section of the wire during the time interval from \( t = 2 \) seconds to \( t = 6 \) seconds, we can follow these steps:

### Step 1: Write down the equation for current
The current \( I \) in the wire is given by the equation:
\[ I = 4 + 2t \]
where \( I \) is in amperes and \( t \) is in seconds.

### Step 2: Relate current to charge
We know that current is the rate of flow of charge, which can be expressed mathematically as:
\[ I = \frac{dq}{dt} \]
This means that the differential charge \( dq \) can be expressed as:
\[ dq = I \, dt \]

### Step 3: Substitute the expression for current
Substituting the expression for current into the equation for \( dq \):
\[ dq = (4 + 2t) \, dt \]

### Step 4: Integrate to find the total charge
To find the total charge \( q \) that has passed through the wire from \( t = 2 \) seconds to \( t = 6 \) seconds, we need to integrate \( dq \) from \( t = 2 \) to \( t = 6 \):
\[ q = \int_{2}^{6} (4 + 2t) \, dt \]

### Step 5: Perform the integration
Now, we can perform the integration:
\[ q = \int (4 + 2t) \, dt = 4t + t^2 + C \]
We will evaluate this from \( t = 2 \) to \( t = 6 \):
\[ q = \left[ 4t + t^2 \right]_{2}^{6} \]

### Step 6: Substitute the limits
Now, we substitute the limits into the integrated function:
\[ q = \left( 4(6) + (6)^2 \right) - \left( 4(2) + (2)^2 \right) \]
Calculating the values:
\[ = (24 + 36) - (8 + 4) \]
\[ = 60 - 12 \]
\[ = 48 \]

### Final Answer
Thus, the quantity of charge that has passed through the cross-section of the wire during the time interval from \( t = 2 \) seconds to \( t = 6 \) seconds is:
\[ \boxed{48 \text{ coulombs}} \]