Question

A square loop of wire of side L= 5.0 cm is in a uniform magnetic field B = 0.16 T. What is the magnetic flux in the loop (a) when the magnetic field B is perpendicular to the face of the loop and (b) when B is at an angle of 30° to the area of the loop? (c) What is the magnitude of the average current in the loop if it has a resistance of 0.012Ω and it is rotated from position (b) to position (a) in 0.14 s?

Answer 1

Side of square=5.0cm

B=0.16T

Magnetic flux$\phi=B.A$

$\phi=BAcos\theta$

A=0.0025m²

$$\phi=0.16\times0.0025cos0°=4\times10^{-4} Wb$$

Part(a)

$\theta=90°$

$\phi=BAcos90=0$

$\phi=0Wb$

Part(b)

$\phi=30°$

$$\phi=BAcos\theta=0.0025\times0.16\times\frac{\sqrt3}{2}$$

$\phi=3.46\times10^{-4}Wb$

Part(c)

$$e=\frac{d\phi}{dt}=A\frac{dB}{dt}=0.0025\times\frac{0.16}{0.14}=2.85\times10^{-3}Wb/sec$$

$i=\frac{e}{R}$

$i=\frac{2.85\times10^{-3}}{0.012}=0.2380A$