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Lady bird [3.3K]

3 years ago

6

10-cm-long wire is pulled along a U-shaped conducting rail in a perpendicularmagnetic field. The total resistance of the wire an

d rail is 0.20 Ω. Pulling the wire with aforce of 1.0 N causes 4.0 W of power to be dissipated in the circuit.
What is the speed of the wire when pulled with a force of 1.0 N?
What is the strength of the magnetic field?

1 answer:

SCORPION-xisa [38]3 years ago

6 0

Answer: Speed = 4m/s

Strength = 7.1T

Explanation: P= 4.0w

N = 1.0

V= ?

Power = force x velocity

Velocity = power/ force

V= 4.0w/1.0N

V = 4m/s

Strength of the magnetic field

B = √PR / lv

Where; l = length 10cm

R= 2.0

P= 4.0w

V= 4m/s

B = √ (4.0w) × (2.0) / (10cm) × (4m/s)

B = 7.1T

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Region II (t = 4 s to t = 6 s)

$v_{6} = -6\,\frac{m}{s} +\int\limits^{6\,s}_{4\,s} {\left(1\,\frac{m}{s^{2}} \right)} \, dt$

$v_{6} = -6\,\frac{m}{s}+\left(1\,\frac{m}{s^{2}} \right) \cdot (6\,s-4\,s)$

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$v_{10} = -4\,\frac{m}{s} +\int\limits^{10\,s}_{6\,s} {\left(2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{10} = -4\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right) \cdot (10\,s-6\,s)$

$v_{10} = 4\,\frac{m}{s}$

Finally, we draw the object's velocity graph as follows. Graphic is attached below.

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