Faraday’s Law of Induction (DP IB Physics: HL): Revision Note

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 the coil 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 40° over a time interval of 0.18 s.

Calculate the magnitude of the average e.m.f induced in the coil.

Answer:

Step 1: Write down the known quantities

• Magnetic flux density, B = 80 mT = 80 × 10^-3 T

• Area, A = 3.5 × 1.4 = (3.5 × 10^-2) × (1.4 × 10^-2) = 4.9 × 10^-4 m²

• Number of turns, N = 350

• Angle of rotation, θ = 40°

• Time interval, Δt = 0.18 s

 Step 2: Write down the equation for Faraday’s law:

$\epsilon =\frac{N(∆\Phi )}{∆t}$

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 B cos θ as it rotates 

• Therefore, the equation to use is:

$N\left(∆\Phi \right)=NBA\cos \theta$

Step 4: Determine the change in magnetic flux linkage

• The coil is initially horizontal, so the plane of the area is parallel to the magnetic field, so ${\theta }_{initial}=90°$

• The initial flux linkage through the coil is:

$N{\Phi }_{initial}=NBA\cos 90°=0$

• The coil rotates through 40°, so the angle between the plane of the area and the field is ${\theta }_{final}=\left(90-40\right)=50°$

• The final flux linkage through the coil is:

$N{\Phi }_{final}=NBA\cos 50°=350\times \left(80\times {10}^{-3}\right)\times \left(4.9\times {10}^{-4}\right)\times \cos 50$

$N{\Phi }_{final}=8.82\times {10}^{-3}\mathrm{Wb}$

• Therefore, the change in flux linkage is:

$N∆\Phi =N\left({\Phi }_{final}-{\Phi }_{initial}\right)=\left(8.82\times {10}^{-3}\right)-0$

$N∆\Phi =8.82\times {10}^{-3}\mathrm{Wb}$

Step 5: Substitute the change in flux linkage and time into Faraday’s law equation:

$\epsilon =\frac{8.82\times {10}^{-3}}{0.18}=0.049\mathrm{V}=49\mathrm{mV}$

The important point to notice is that an emf is induced in a conductor in a magnetic field if there is change in flux linkage. This means the conductor (e.g. a coil) must cut through the field lines to have an emf (and hence a current) induced.