What Happens to di/dt When a Switch in an Inductor Circuit Opens?

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In a circuit with a 10V battery, a 1 henry inductor, and a switch, when the switch opens, the inductor's behavior must be analyzed using the equation V = L x di/dt. The initial current is 1mA, and while the open switch creates a high impedance, it does not lead to infinite voltage; instead, the applied voltage across the inductor becomes zero. To find di/dt, one must set up the circuit equations considering the potential drops and solve the resulting differential equation. The initial current value is crucial for determining the current as a function of time. Understanding these concepts is essential for solving the problem accurately.
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Homework Statement



A circuit has a 10v battery, an 1 henry inductor, and a switch connected in series. The circuit's total resistance is 10k ohms. If the inductors maximum current is 1mA and the switch is suddenly opened how do I calculate di/dt?

Wouldn't the open switch be an infinite impedance and produce infinite voltage across the inductor?


Homework Equations



V=L x di/dt

The Attempt at a Solution



Not sure where to even start?
 
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first consider the switch is closed and current is flowing at steady rate(think about it, why it should flow at a steady rate? if u don't know it already), the value is given as 1mA. u can also calculate it from given data (how?), had it not given.
now consider the switch is suddenly open. the physical consideration should be, the applied voltage is zero (and not the resistance is infinity. why?).
now set up the ckt eqn. considering potential drops across the inductor and the resistor and applied voltage zero. solve the differential eqn (u will need the value i initial = 1 mA here). and find i as a function of t. put the given values for inductor and resistor and find di/dt by differentiation.
 
Can you provide a drawing of the circuit? The problem seems a bit strange for an intro physics class.
 
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