Answer :
a) the flash of light lasts for 40 ms. b) the current flows clockwise as the loop exits the field. c) the power dissipated in the bulb during the flash is 3.06 W.
Explanation:
a. The time duration of the flash of light can be calculated using the formula:
Δt = L/ v
where L is the perimeter of the loop and v is the velocity of the loop. The perimeter of the loop is:
L = 2(length + width) = 2(40 cm + 10 cm) = 100 cm = 1 m
Converting the velocity to m/s, we have:
v = 25 m/s
Therefore, the time duration of the flash is:
Δt = L/v = 1 m / 25 m/s = 0.04 s = 40 ms
So, the flash of light lasts for 40 ms.
b. The direction of the current flow can be determined using Lenz's law. According to Lenz's law, the direction of the induced current in a circuit is such that it opposes the change in magnetic flux that produced it.
As the loop is pulled out of the magnetic field, the flux through the loop decreases. To oppose this decrease, the induced current should produce a magnetic field in the opposite direction to that of the external field. By the right-hand rule, this means the current should flow in a clockwise direction when viewed from above the loop.
So, the current flows clockwise as the loop exits the field.
c. The power dissipated in the bulb can be calculated using the formula:
P = I^2R
where I is the current flowing through the loop and R is the resistance of the bulb. The resistance of the bulb is given as 25 ohms.
To find the current, we can use Faraday's law of electromagnetic induction, which states that the voltage induced in a circuit is equal to the rate of change of magnetic flux through the circuit. The rate of change of flux through the loop can be calculated using:
dΦ/dt = B(dA/dt)
where B is the magnetic field, A is the area of the loop, and dA/dt is the rate of change of area (which is equal to the velocity v of the loop as it exits the field).
The area of the loop is:
A = length x width = 40 cm x 10 cm = 400 cm^2 = 0.04 m^2
Converting the velocity to m/s, we have:
v = 25 m/s
So, the rate of change of area is:
dA/dt = -v x width = -25 m/s x 0.1 m = -2.5 m^2/s
Therefore, the rate of change of flux is:
dΦ/dt = B(dA/dt) = 3.5 T x (-2.5 m^2/s) = -8.75 Wb/s
The voltage induced in the circuit is equal to the rate of change of flux multiplied by the number of turns in the loop. Since there is only one turn in the loop, the induced voltage is:
V = -dΦ/dt = 8.75 V
The current flowing through the loop is:
I = V/R = 8.75 V / 25 ohms = 0.35 A
Finally, the power dissipated in the bulb is:
P = I^2R = (0.35 A)^2 x 25 ohms = 3.06 W
So, the power dissipated in the bulb during the flash is 3.06 W.
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