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19 days ago

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

Physics

2 answers:

Ostrovityanka [2.2K]19 days ago

8 0

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

Explanation:

According to Faraday's Law, the magnitude of the induced electromotive force (emf) in the coil corresponds to the rate of flux change that links through the coil:

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

where

N = 505 represents the number of turns in the coil

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

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

$\epsilon = 0.166 V$

Rotating the coil from perpendicular to parallel alignment with the Earth's magnetic field results in the final flux equaling zero, making the magnitude of flux change simply the initial flux:

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

where

B denotes the intensity of the magnetic field

A signifies the area of the coil

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

The area of the coil can be expressed as

$A=\pi r^2$

where

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

By substituting all values into equation (1) and solving for B, we obtain:

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

For further reading on magnetic fields:

Softa [2K]19 days ago

7 0

Answer:

Earth's magnetic field strength is measured at 4.825 x 10⁻⁵ T

Explanation:

Provided details;

Turns count; N = 505 turns

Diameter of the circular loop, d = 15.5 cm

Average emf, V = 0.166 V

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

Finally, after the coil is rotated 90.0°, $\theta_f = 90^o$

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

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

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