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Answer :
Final answer:
The flash of light lasts for 16 ms after the loop is pulled from the magnetic field. The current flows in the counter-clockwise direction as the loop exits the field due to Lenz's Law. The power dissipated in the bulb during the flash is 576W.
Explanation:
To determine the duration of the flash of light, we need to calculate the time it takes for the loop to exit the magnetic field. The formula to calculate this time is t = L/v, where L is the length of the loop and v is the velocity. Plugging in the values, we have t = 0.40 m / 25 m/s = 0.016 s = 16 ms.
As the loop exits the magnetic field, the current flows in the counter-clockwise direction. This can be determined using Lenz's Law, which states that the induced current will always oppose the change in magnetic flux. So, as the loop is pulled out of the magnetic field, the induced current will create a magnetic field that opposes the external field, and this requires the current to flow in the counter-clockwise direction.
To calculate the power dissipated in the bulb during the flash, we can use the formula P = I²R, where P is power, I is current, and R is resistance. Since we know the resistance is 25 ohms and the power is dissipated in the bulb, we just need to find the current flowing in the circuit. The current can be calculated using Ohm's Law: I = V/R, where V is the voltage. Assuming a standard voltage of 120V, we have I = 120V / 25 ohms = 4.8A. Plugging this value into the power formula, we get P = (4.8A)² * 25 ohms = 576W.
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a) The light flashes as the circuit leaves the field at a speed of 16 ms.
b) The current flow as the loop exits the field in the clockwise direction.
c) The power dissipated in the bulb during the flash is 0.04 W.
To reply to these questions, we will utilize Faraday's Law, which states that a changing attractive field actuates an electromotive drive (EMF) in a circuit, and the initiated EMF is rise to the rate of alter of attractive flux through the circuit.
a) The attractive flux through the circle is given by the item of the attractive field, region of the circle, and cosine of the point between the attractive field and the ordinary to the plane of the circle.
As the circle is pulled out of the attractive field, the magnetic flux through the circle diminishes, and thus, an EMF is actuated within the circle. This initiated EMF drives a current through the light bulb, causing it to light up.
The time term of the streak of light can be decided from the time taken by the circle to move out of the attractive field.
The removal voyage by the circle is 40 cm, and the speed is 25 m/s, so the time taken is:
t = d/v = 0.4 m / 25 m/s = 0.016 s = 16 ms
Subsequently, the streak of light endures for 16 ms.
b) Concurring to Lenz's Law, the course of the initiated current is such that it contradicts the alter within the attractive flux that produces it. As the circle is pulled out of the attractive field, the attractive flux through the circle diminishes.
Hence, the actuated current flows in a course that makes a magnetic field that restricts the initial attractive field. This could be accomplished by the induced current streaming clockwise as seen from above. Hence, the reply is clockwise.
c) The control scattered within the light bulb can be calculated utilizing the equation P = V²/R, where V is the voltage over the bulb and R is its resistance.
The voltage over the bulb is break even with to the initiated EMF, which can be calculated from Faraday's Law. The attractive flux through the circle changes at a rate of (40 cm) x (25 m/s) = 1 T.m²/s.
The region of the circle is (40 cm) x (10 cm) = 0.04 m². The cosine of the point between the attractive field and the ordinary plane of the circle is 1 (since the circle is opposite to the field). Subsequently, the induced EMF is:
EMF = -d(phi)/dt = -NA(dB/dt)
= -(1)(0.04 m²)(1 T.m²/s)/0.016 s
= -1 V
The negative sign indicates that the actuated EMF is within the inverse course of the current stream. Subsequently, the voltage over the light bulb is:
V = -EMF = 1 V
The power dissipated within the bulb is:
P = V²/R = (1 V)²/25 ohm = 0.04 W
Subsequently, the control scattered within the bulb during the streak is 0.04 W.
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