A long wire lying along the x-axis carries a current of 3.20 A in the +x directi

Question

A long wire lying along the x-axis carries a current of 3.20 A in the +x direction. There is a uniform magnetic field present, given by B=0.003 i + 0.004 j + 0.002 k, where i,j,k are the unit vectors along the cartesian coordinate axes in units of Tesla. Calculate the y-component of the magnetic force acting on a segment of wire of length L = 21.0 cm. -0.00168 N Use the formula for the magnetic force on a force on a straight wire carrying current in a uniform magnetic field. You can find the cross product by applying the right hand rule separately for each B component.

Explanation / Answer

magnetic force:

F=L(i*B) =0.21*(3.20i*{0.003i+0.004j+0.002k})

F=0.21(0 i+ 0.0128 j + 0.0064 k)

F= 0 i+0.00268 j + 0.001344 k

Y component of force: 0.00268

Magnitude of force:

F=(.00268²+.001344²) = 0.00299 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.