Question

The magnetic flux through a coil of resistance 5 \(\Omega\) increases with time as: \(\Phi = (2.0 t^3 + 5.0 t^2 + 6.0 t)\) mWb. Find the magnitude of induced current through the coil at \(t = 2\) s.

Hint:

When calculating induced current due to changing magnetic flux, always check the units of magnetic flux and convert them to standard units (webers) if necessary to ensure the correct calculation of induced emf using Faraday's law.

Answer 1

The induced emf (\( \mathcal{E} \)) in the coil is given by Faraday's law of induction, which states that: \[ \mathcal{E} = - \frac{d\phi}{dt} \] where \( \phi \) is the magnetic flux. Substituting the given expression for \( \phi \), we first find the derivative of \( \phi \) with respect to time \( t \). The given flux is: \[ \phi = 2.0 t^3 + 5.0 t^2 + 6.0 t \, mWb} \] Now, differentiate \( \phi \) with respect to \( t \): \[ \frac{d\phi}{dt} = 3 \times 2.0 t^2 + 2 \times 5.0 t + 6.0 = 6.0 t^2 + 10.0 t + 6.0 \] At \( t = 2 \, s} \), substitute the value of \( t \) into the derivative: \[ \frac{d\phi}{dt} = 6.0 \times (2)^2 + 10.0 \times (2) + 6.0 = 6.0 \times 4 + 10.0 \times 2 + 6.0 = 24 + 20 + 6 = 50.0 \, mWb/s} \] Thus, the induced emf is: \[ \mathcal{E} = - 50.0 \, mV} \] Now, using Ohm's law, the induced current \( I \) through the coil is given by: \[ I = \frac{\mathcal{E}}{R} \] where \( R = 5 \, \Omega \) is the resistance of the coil. Substituting the values: \[ I = \frac{50.0 \times 10^{-3}}{5.0} = 10.0 \times 10^{-3} = 0.01 \, A} = 10 \, mA} \] Thus, the magnitude of the induced current at \( t = 2 \, s} \) is \( 10 \, mA} \).