# Calculating Induced E.m.f ## Faraday's Law with Lenz's Law - Combining Lenz's Law into Faraday's Law gives the complete equation for induced e.m.f: $$\varepsilon = -\frac{\Delta(N\Phi)}{\Delta t}$$ - Where: - $\varepsilon$ = induced e.m.f (V) - $\Delta(N\Phi)$ = change in flux linkage (Wb turns) - $\Delta t$ = time interval (s)

The negative sign in the equation represents Lenz's Law - the induced e.m.f opposes the change causing it

- The negative sign shows the induced e.m.f $\varepsilon$ is set up in an 'opposite direction' to oppose the changing flux linkage - This equation reveals that the **gradient** of a graph of magnetic flux linkage against time represents the **magnitude** of the induced e.m.f - Note: if the gradient is positive, the induced e.m.f is negative - This is again due to Lenz's law - the e.m.f opposes the effects of the changing flux linkage ## Worked Example: Static Coil Rotation **Question:** A small rectangular coil contains 350 turns of wire. The longer sides are 3.5 cm and the shorter sides are 1.4 cm. The coil is held between the poles of a large magnet so that it can rotate about an axis through its centre. The magnet produces a uniform magnetic field of flux density 80 mT between its poles. The coil is positioned horizontally and then turned through an angle of 90° in a time of 0.18 s. Calculate the magnitude of the average e.m.f induced in the coil. **Step 1: Write down the known quantities** - Magnetic flux density, $B = 80 \text{ mT} = 80 \times 10^{-3} \text{ T}$ - Area, $A = 3.5 \times 1.4 = (3.5 \times 10^{-2}) \times (1.4 \times 10^{-2}) = 4.9 \times 10^{-4} \text{ m}^2$ - Number of turns, $N = 350$ - Time interval, $\Delta t = 0.18 \text{ s}$ **Step 2: Write out the equation for Faraday's law** $$\varepsilon = \frac{\Delta(N\Phi)}{\Delta t}$$ **Step 3: Write out the equation for the change in flux linkage** - The number of turns $N$ and the coil area $A$ stay constant - The flux through the coil changes as it rotates - Therefore, the change in flux linkage can be written as: $$\Delta(N\Phi) = NA(\Delta B)$$ **Step 4: Determine the change in magnetic flux linkage** - The initial flux through the coil is zero (flux lines are parallel to the coil face) - The final flux through the coil is 80 mT (flux lines are perpendicular to the coil face) - This is because the coil begins horizontally in the field and is rotated 90° - Therefore, the change in flux linkage is: $$\Delta(N\Phi) = 350 \times (4.9 \times 10^{-4}) \times (80 \times 10^{-3}) = 0.014 \text{ Wb turns}$$ **Step 5: Substitute change in flux linkage and time into Faraday's law equation** $$\varepsilon = \frac{0.014}{0.18} = 0.076 \text{ V}$$ --- ## E.M.F Induced in a Rotating Coil - When a coil rotates in a uniform magnetic field, the flux through the coil varies as it rotates - Since e.m.f is the rate of change of flux linkage, this means the e.m.f will also change as it rotates - The maximum e.m.f occurs when the coil **cuts through** the most field lines - The e.m.f induced is an **alternating voltage** ### Maximum and Minimum E.M.F

Maximum e.m.f when $\theta = 90°$ - field lines parallel to the plane of the coil

- The maximum e.m.f is when the coil cuts through the field lines when they are **parallel** to the plane of the coil - This means that the e.m.f is: - **Maximum** when $\theta = 90°$ - magnetic field lines are parallel to the plane of the area (or the normal to the area is perpendicular to the field lines) - **Zero** when $\theta = 0°$ - magnetic fields lines are perpendicular to the plane of the area (or the normal to the area is parallel to the field lines) - This is the **opposite** of the maximum and minimum flux through the coil ### Flux Linkage in a Rotating Coil - The flux linkage can also be written as: $$N\Phi = BAN\cos(\theta)$$ - Where the angle $\theta$ depends on the angular speed of the coil, $\omega$: $$\theta = \omega t$$ ### E.M.F Equation for Rotating Coil - The induced e.m.f, $\varepsilon$ from Faraday's Law depends on the rate of change of flux linkage - This means it can also be written as: $$\varepsilon = BAN\omega \sin(\theta)$$ - Where: - $\varepsilon$ = e.m.f (V) - $B$ = magnetic flux density (T) - $A$ = cross-sectional area (m²) - $N$ = number of turns - $\omega$ = angular velocity of the rotating coil (rad s⁻¹) - $\theta$ = angle between the magnetic field, $B$ and the normal to the area, $A$ (rads) - The equation shows that the e.m.f varies **sinusoidally** and it is **90° out of phase** with the flux linkage - The graph below shows how both flux linkage and e.m.f vary as the coil rotates through 360° (values are normalized):

E.m.f and flux linkage are 90° out of phase - when one is maximum, the other is zero

### Key Relationships Summary | Coil Orientation | Flux Linkage | E.M.F | |------------------|-------------|-------| | Plane perpendicular to field lines ($\theta = 0°$) | Maximum ($BAN$) | Zero | | Plane parallel to field lines ($\theta = 90°$) | Zero | Maximum ($BAN\omega$) | ## Worked Example: Rotating Coil Maximum E.M.F **Question:** A rectangular coil with 40 turns, each with an area of 0.5 m², is rotated at 42 rad s⁻¹ in a uniform 3.15 mT magnetic field. Calculate the maximum e.m.f induced in the coil. **Step 1: Write the known quantities** - Number of turns, $N = 40$ - Area, $A = 0.5 \text{ m}^2$ - Angular velocity, $\omega = 42 \text{ rad s}^{-1}$ - Magnetic flux density, $B = 3.15 \text{ mT} = 3.15 \times 10^{-3} \text{ T}$ **Step 2: Write down the e.m.f equation** $$\varepsilon = BAN\omega \sin(\omega t)$$ **Step 3: Determine when the maximum e.m.f will be** - The maximum e.m.f occurs when $\sin(\omega t) = \pm 1$ - Or when the coil is parallel to the magnetic field **Step 4: Substitute in the values** $$\varepsilon = (3.15 \times 10^{-3}) \times 0.5 \times 40 \times 42 \times (\pm 1) = \pm 2.6 \text{ V}$$ --- ## Key Points to Remember ### When is E.M.F or Flux Linkage Maximum? **When the plane of the coil is perpendicular to the field lines:** - The flux linkage is at its **maximum** - The e.m.f = **0** **When the plane of the coil is parallel to the field lines:** - The flux linkage is **0** - The e.m.f is at its **maximum** ### Calculator Mode - Since $\omega$ is in units of rad s⁻¹, make sure your calculator is in **radians mode** before doing calculations - This applies to all trigonometric functions involving angles in the equations ### Magnitude of E.M.F - The 'magnitude' of the e.m.f just means its size, rather than its direction - This is often what is required in exam questions, so the minus sign in Lenz's law is not necessarily required in calculations - However, you may be expected to explain the significance of the negative sign in the equation - Be prepared to interpret it as an expression of Lenz's Law ### Understanding the Delta Symbol - The Greek letter delta, '$\Delta$', simply means 'change in' - So $\Delta(N\Phi)$ means the change in flux linkage - And $\Delta t$ means the change in time (or time interval)