Question

Question 3. A wire 25.0cm long lies along the z-axis and carries a current of 9.00A in the +z-direction. The a magnetic field is uniform and has components Bx = -0.242T, By= -0.985, and B2=-0.336. a. Find the components of the magnetic force on the wire? b. What is the magnitude of the net magnetic force on the wire? ​

Answer 1

a.

The components of the force are Fx = 2.2163 N, Fy = -0.5445 N and Fz = 0 N

The force on a current carrying conductor in a magnetic field is given by F = iL × B where i = current = 9.00 A, L = 25.0 cmk = 0.25 mk (since the conductor is along the z-direction). B = magnetic field. Since B has component Bx = -0.242T, By= -0.985, and Bz = -0.336, B = -0.242i + (-0.985j) + (-0.336)k = -0.242i - 0.985j - 0.336)k.

So, F = iL × B

F = 9.00 A{(0.25 m)k × [-0.242Ti + (-0.985Tj) + (-0.336T)k]T}

F = 9.00 A{(0.25 m)k × (-0.242T)i + (0.25 m)k × (-0.985Tj) + (0.25 m)k × (-0.336T)k]}

F = 9.00 A{-0.0605mT)k × i + (-0.24625 mT)k × j + (-0.084 m)k × k]}

F = 9.00 A{-0.0605mT)j + (-0.24625 mT) × -i + (-0.084 mT) × 0]}

F = 9.00 A{-0.0605mT)j + (0.24625 mT)i + 0 mT]}

F = -0.5445 AmT)j + (2.21625 AmT)i + 0 AmT]}

F = -0.5445j + 2.21625i + 0 k

F = (2.2163i - 0.5445j + 0 k) N

So, the components of the force are Fx = 2.2163 N, Fy = -0.5445 N and Fz = 0 N

b.

The magnitude of the net force on the wire is 2.282 N

The net force F = √(Fx² + Fy² + Fz²)

F = √[(2.2163 N)² + (-0.5445 N)² + (0 N)²)

F = √[(4.912 N)² + 0.2964 N)² + (0 N)²)

F = √[5.2084 N)²

F = 2.2822 N

F ≅ 2.282 N

So, the magnitude of the net force on the wire is 2.282 N

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