Question #101119

1. A rectangular coil measuring 200 mm by 100 mm is mounted such

that it can be rotated about the midpoints of the 100 mm sides.

The axis of rotation is at right angles to a magnetic field of uniform

flux density 0.05 T. Calculate the flux in the coil for the following

conditions:

(a) the maximum flux through the coil and the position at which

it occurs;

(b) the flux through the coil when the 100 mm sides are inclined

at 45° to the direction of the flux


2. Three resistors of 6 Ω, 9 Ω and 15 Ω are connected in

parallel to a 9 V supply. Calculate: (a) the current in

each branch of the network; (b) the supply current;

(c) the total effective resistance of the network.

Expert's answer

1. Magnetic flux through a flat coil Φ=BAcos⁡α\Phi = B A \cos\alpha ,

where:

B - flux density, TA=w∗h - area of the coil, m2α - angle between the magnetic induction vector  and the normal to the plane of the coil w,h - width and height of the coil, m \quad B \text{ - flux density, T} \\ \quad A = w * h \text{ - area of the coil, }m^2 \\ \quad \alpha \text{ - angle between the magnetic induction vector } \\ \text{ and the normal to the plane of the coil } \\ \quad w, h \text{ - width and height of the coil, m }


1.a The maximum flux through the coil occurs when the plane of the coil is perpendicular to the magnetic field.

Φmax=0.05∗0.2∗0.1∗cos⁡0=10−3 (Wb)\quad \Phi_{max} = 0.05 * 0.2 * 0.1 * \cos 0 = 10^{-3} \text{ (Wb)}


1.b

Φ45∘=0.05∗0.2∗0.1∗cos⁡45∘≈7.07∗10−4 (Wb)\quad \Phi_{45^\circ} = 0.05 * 0.2 * 0.1 * \cos {45^\circ} \approx 7.07 * 10^{-4} \text{ (Wb)}



2. Will use Ohm's law I=VRI = \cfrac V R


2.a Current in each branch of the network

I6Ω=9/6=1.5 (A)I9Ω=9/9=1 (A)I15Ω=9/15=0.6 (A)\quad I_{6\Omega} = 9 / 6 = 1.5 \text{ (A)} \\ \quad I_{9\Omega} = 9 / 9 = 1 \text{ (A)} \\ \quad I_{15\Omega} = 9 / 15 = 0.6 \text{ (A)}


2.b Supply current

IΣ=I6Ω+I9Ω+I15Ω=1.5+1+0.6=3.1 (A)\quad I_\Sigma = I_{6\Omega} + I_{9\Omega} + I_{15\Omega} = 1.5 + 1 + 0.6 = 3.1 \text{ (A)}


2.c Total effective resistance of the network

RΣ=VIΣ=9/3.1≈2.9 (Ohm)\quad R_\Sigma = \cfrac V I_\Sigma = 9 / 3.1 \approx 2.9 \text{ (Ohm)}



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