Question

4. A 505-turn circular-loop coil with a diameter of 15.5 cm is initially aligned so that

Answer 1

The strength of the magnetic field is $4.8\cdot 10^{-5} T$

Explanation:

According to Faraday's Law, the magnitude of the induced emf in the coil is equal to the rate of changeof the flux linkage through the coil:

$\epsilon = \frac{N\Delta \Phi}{\Delta t}$ (1)

where

N = 505 is the number of turns in the coil

$\Delta \Phi$ is the change in magnetic flux through the coil

$\Delta t = 2.77 ms = 2.77\cdot 10^{-3} s$ is the time interval

$\epsilon = 0.166 V$

The coil is rotated from a position perpendicular to the Earth's magnetic field to a position parallel to it, so the final flux is zero, and the magnitude of the flux change is simply equal to the initial flux:

$\Delta \Phi = B A cos \theta$

where

B is the strength of the magnetic field

A is the area of the coil

$\theta=0^{\circ}$ is the angle between the normal to the coil and the field

The area of the coil can be written as

$A=\pi r^2$

where

$r=\frac{15.5 cm}{2}=7.75 cm = 7.75\cdot 10^{-2} m$ is its radius

Substituting everything into eq.(1) and solving for B, we find:

$\epsilon= \frac{NB\pi r^2 cos \theta}{\Delta t}\\B=\frac{\epsilon \Delta t}{\pi r^2 cos \theta}=\frac{(0.166)(2.77\cdot 10^{-3})}{(505)\pi (7.75\cdot 10^{-2})^2(cos 0^{\circ})}=4.8\cdot 10^{-5} T$

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Answer 2

Answer:

The value of Earth's magnetic field is 4.825 x 10⁻⁵ T

Explanation:

Given;

Number of turns; N = 505-turn

Diameter of the circular-loop, d = 15.5 cm

Average emf, V = 0.166 V

change in time, t = 2.77 ms

$V_{avg} =- N\frac{d \phi}{dt} -------equation (i)\\\\\phi = BACos \theta \\\\d \phi = BACos \theta_f - BACos \theta_i\\\\V_{avg.} = -N(\frac{BACos \theta_f - BACos \theta_i}{dt})\\\\V_{avg.} = N(\frac{BACos \theta_i - BACos \theta_f}{dt}) ---------equation(ii)$

initially, when the plane of the circular loop is perpendicular to Earth's magnetic field, $\theta _i = 0^o$

Finally, when the coil was rotated 90.0°, $\theta_f = 90^o$

$V_{avg.} = NBA(\frac{Cos 0 -Cos 90}{t} )\\\\V_{avg.} = \frac{NBA}{t} \\\\B = \frac{V_{avg.}*t}{NA} -------------equation(iii)\\\\But, A = \frac{\pi d^2}{4} \\\\B = \frac{4*V_{avg.}*t}{N*\pi d^2}\\\\substitute \ the \ given \ values\\\\B = \frac{4*(0.166)*(2.77*10^{-3})}{(505)*\pi* (0.155)^2} = 4.825*10^{-5} \ T$

Thus, the value of Earth's magnetic field is 4.825 x 10⁻⁵ T