A coil 3.10 cm in radius, containing 410 turns, is placed in a uniform magnetic field that varies with time according to B = (0.0170 T/s)t + (2.60 x 10-5 T/s4)t4. The coil is connected to a 6.10 Ω resistor, and its plane is perpendicular to the magnetic field. You can ignore the resistance of the coil. What is the magnitude of the current in the resistor at time t = 5.90 s? Give your answer in unit of milliampere (mA).

So, I have calculated the field at t=5.9. My question is, how do I get indued current from the field? I have no idea where to start, other than knowing that a varying magnetic field induces a current. I cannot seem to find any formulas in the course textbook pertaining to induced current from a magnetic field, let alone a coil with a resistor.

Please help me!!!


1 Answer 1


I will try to give you the solution without the numerical application (but it is really simple to do the numerical computations).

Using Fraday's induction formula the EMF :

$$ e = -\frac{d\phi}{dt} = -\left(\frac{\partial{\phi}}{\partial{t}}\right)_{v=0}-\left(\frac{\partial{\phi}}{\partial{t}}\right)_{B=constante}$$

Now since your coil is fixed relative to the inducing magnetic field system and your magnetic field is varying, $e$ becomes :

$$ e = -\frac{\partial{(\int_{S}{\vec B.\vec ds})}}{\partial{t}}$$

$$\int_{S}{\vec B.\vec ds} = \pi r^2\times B$$

and the induced curent i : $$i = \frac{e}{R + iX} = \frac{\pi r^2 \dot{B}(5.9)}{R + iX} = \frac{e}{6.1} $$

The $X$ is the some of the inductance and capacitance (you don't have them ) and $R$ is the some of the coil and resistor resistances.

I hope this helped


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