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 horizontal 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, ∆t = 0.18 s

Step 2: Write down the equation for the magnitude of the induced e.m.f. from 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 coil is parallel to the magnetic field and its normal is perpendicular to the 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 field and the normal to the plane of the coil 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 magnetic flux linkage through the conductor must change, for example by cutting field lines or by the field itself changing.