(a) The current through a wire is a steady 2.5 amps. How much current passes through it between t = 0 seconds and t = 4 × 10-4 seconds?

(b) The current through a wire is given by I(t) = I0e -at, where I0 = 2.5 amps and a = 6 × 103 s-1. How much electric charge passes through the wire between t = 0 seconds and t = 4 × 10-4 seconds?

To find the amount of current passing through a wire between two given times, we need to calculate the total charge. The formula to calculate the charge is:

Q = I × t

where Q is the charge in Coulombs (C), I is the current in Amperes (A), and t is the time in seconds (s).

(a) Given that the current is a steady 2.5 amps, and the time interval is from t = 0 seconds to t = 4 × 10^−4 seconds, we can substitute these values into the formula:

Q = I × t
= 2.5 A × (4 × 10^−4 s)

First, let's calculate 4 × 10^−4:

4 × 10^−4 = 0.0004

Now, we can substitute this value into the equation:

Q = 2.5 A × 0.0004 s
= 0.001 Coulombs

Therefore, the amount of charge passing through the wire between t = 0 seconds and t = 4 × 10^−4 seconds is 0.001 Coulombs.

(b) In this case, we have an exponential function I(t) = I0e^(-at), where I0 = 2.5 amps and a = 6 × 10^3 s^−1.

To find the amount of charge passing through the wire between t = 0 seconds and t = 4 × 10^−4 seconds, we need to integrate the current function over the given time interval:

Q = ∫[0 to 4 × 10^−4] I(t) dt

To integrate the exponential function, we can use the power rule for integration:

∫ e^(-ax) dx = - (1/a) e^(-ax) + C

where C is the constant of integration.

Applying the power rule to our current function I(t) = I0e^(-at), we get:

∫ I0e^(-at) dt = - (1/a) I0e^(-at) + C

Now, we can substitute the limits of integration t = 0 and t = 4 × 10^−4 into the equation:

Q = [- (1/a) I0e^(-at)] [0 to 4 × 10^−4]

To simplify the equation further, let's substitute the given values I0 = 2.5 amps and a = 6 × 10^3 s^−1:

Q = [- (1/a) I0e^(-at)] [0 to 4 × 10^−4]
= [- (1/(6 × 10^3 s^−1)) (2.5 A) e^(-(6 × 10^3 s^−1) t)] [0 to 4 × 10^−4]

After evaluating the expression at the lower and upper limits, we can find the amount of charge passing through the wire between t = 0 seconds and t = 4 × 10^−4 seconds.