A Nichrome wire 44 cm long and 0.30 mm in diameter is connected to a 3.1 V flashlight battery. What is the electric field inside
1 answer:
Answer:
7.05 Volts/m
Explanation:
L = length of the Nichrome wire = 44 cm = 0.44 m
V = Potential difference across the end of the wire = battery voltage = 3.1 Volts
E = magnitude of electric field inside the wire
Magnitude of electric field inside the wire is given as
Inserting the values
E = 7.05 Volts/m
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Answer:
Work done, W = 0.0219 J
Explanation:
Given that,
Force constant of the spring, k = 290 N/m
Compression in the spring, x = 12.3 mm = 0.0123 m
We need to find the work done to compress a spring. The work done in this way is given by :
W = 0.0219 J
So, the work done by the spring is 0.0219 joules. Hence, this is the required solution.
Answer:
Please refer to the figure.
Explanation:
The crucial point here is to calculate the enclosed current. If the current I is flowing through the whole cross-sectional area of the wire, the current density is
The current density is constant for different parts of the wire. This idea is similar to that of the density of a glass of water is equal to the density of a whole bucket of water.
So,
This enclosed current is now to be used in Ampere’s Law.
Here, represents the circular path of radius r. So we can replace the integral with the circumference of the path,
.
As a result, the magnetic field is
