Question

A rectangular coil of wire is situated in a constant magnetic field whose magnitude is 0.50T. The coil has an area of $2m^2$. Determine the magnetic flux for the three orientations, $\theta =0^{°},{60}^{°}$ and ${90}^{°}$. Where is the angle between area vector and magnetic field intensity?

Answer 1

Step 1: Given data

Magnetic field intensity, $\overrightarrow{B}=0.50T$

Area of the coil, $\overrightarrow{A}=2m^2$

The angle between area vector and magnetic field intensity, $\theta =0^{°},{60}^{°}$and ${90}^{°}$

Magnetic flux through a surface, $\varphi =\overrightarrow{B}.\overrightarrow{A}=BA\cos \theta$

Step 2: Calculation of the magnetic flux

For $\theta =0^{°}$, magnetic flux, $\varphi =BA\cos \theta$

$=0.5\times 2\times \cos 0=0.5\times 2\times 1$

$=1Wb$

For$\theta =0^{°}$, the magnetic flux is $1Wb$

For $\theta ={60}^{°}$, magnetic flux, $\varphi =BA\cos \theta$

$=0.5\times 2\times \cos 60=0.5\times 2\times \frac{1}{2}$

$=0.5Wb$

For $\theta ={60}^{°}$, the magnetic flux is $0.5Wb$.

For $\theta ={90}^{°}$, magnetic flux, $\varphi =BA\cos \theta$

$=0.5\times 2\times \cos 90=0.5\times 2\times 0$

$=0$

For $\theta ={90}^{°}$, the magnetic flux is $0$.

Step 3: Final answer

Thus the magnetic flux for the three orientations $\theta =0^{°},{60}^{°}$ and ${90}^{°}$are 1Wb, 0.5Wb, and 0 respectively.