Respuesta :
Answer:
a. 169.4 N b. 65.08 m/s
Explanation:
Here is the complete question
A device called a railgun uses the magnetic force on currents to launch projectiles at very high speeds. An idealized model of a railgun is illustrated in the figure. A 1.2 V power supply is connected to two conducting rails. A segment of copper wire, in a region of uniform magnetic field, slides freely on the rails. The wire has a 0.85 mΩ resistance and a mass of 5.2 g . Ignore the resistance of the rails.
a. What is the magnitude of the force on the wire?
b. What will be the wire's speed after it has slid a distance of 6.5cm ?
Solution
The force on a current carrying conductor in a magnetic field is given by F = BIL where B = magnetic field strength = 0.8 T, I = current = V/R . V = voltage = 1.2 V and R = resistance = 0.85 × 10⁻³ Ω and L = length of conductor = 15 cm = 0.15 m.
F = BIL = BVL/R = 0.8 × 1.2/0.85 × 10⁻³ × 0.15 N = 0.144/0.85 × 10⁻³ N = 169.4 N
b. First we find its acceleration from F = ma, a = F/m
Since the copper wire of mass m = 5.2 g = 5.2 × 10⁻³ starts from rest and moves a distance d = 6.5 cm = 0.065 m,
we use v² = u² + 2ad, u = 0 ⇒ v² = 2ad ⇒ v = √(2ad) = √(2Fd/m) = √(2 × 169.4 × 0.065/5.2 × 10⁻³) m/s = √4235 = 65.08 m/s
a)The magnitude of the force on the wire will be 169.4 N
b)The wire's speed after it has slid a distance of 6.5cm will be 65.08 m/s.
What is speed?
Speed is defined as the rate of change of the distance or the height attained. it is a time-based quantity. it is denoted by u for the initial speed while u for the final speed. its si unit is m/sec.
The given data in the problem is;
B is the magnetic field strength = 0.8 T
L is the length of the wire = 0.85 m
m is the mass = 5.2 g
v is the speed of wire=?
V is the voltage = 1.2 V
u is the initial speed = 0 m/sec
The magnetic force on a current-carrying conductor in a magnetic field is found as;
[tex]\rm F=BIL \\\\ \rm F =\frac{BVL}{R} \\\\ \rm F =\frac{0.8 \times 1.2 \times 0.15}{0.85 \times 10^{-3}} \\\\ \rm F=169.4 N[/tex]
Hence the magnitude of the force on the wire will be 169.4 N
b)The wire's speed after it has slid a distance of 6.5cm will be 65.08 m/s.
From Newton's second law;
[tex]\rm v^2=u^2+2ad \\\\ v^2=2ad \\\\ v=\sqrt{\frac{2Fd}{m} } \\\\ v=\sqrt{\frac{2\times 169.4 \times 0.065}{5.2 \times 10^{-3}} } \\\\ \rm v=65.08 \m/sec[/tex]
Hence the wire's speed after it has slide a distance of 6.5cm will be 65.08 m/s.
To learn more about the speed refer to the link;
https://brainly.com/question/7359669
