A wire, of length L = 3. 8 mm, on a circuit board carries a current of I = 2. 54 μA in the j direction. A nearby circuit element generates a magnetic field in the vicinity of the wire of B = Bxi + Byj + Bzk, where Bx = 6. 9 G, By = 2. 6 G, and Bz = 1. 1 G. A) Calculate the i component of the magnetic force Fx, in newtons, exerted on the wire by the magnetic field due to the circuit element.

B) Calculate the k component of the magnetic force Fz, in newtons, exerted on the wire by the magnetic field due to the circuit element.

C) Calculate the magnitude of the magnetic force F, in newtons, exerted on the wire by the magnetic field due to the circuit element

Answers

Answer 1

The i component of the magnetic force on the wire is 1.06 × 10^-13 N. The k component of the magnetic force on the wire is 6.69 × 10^-14 N. The magnitude of the magnetic force on the wire is 1.26 × 10^-13 N.

To calculate the i component of the magnetic force, we use the formula:

F = I * L x B

where I is the current, L is the length of the wire, B is the magnetic field, and x represents the cross product.

The cross product of L and B gives a vector perpendicular to both L and B, which is in the i direction. So we only need to find the magnitude of the cross product and multiply it by I to get Fx.

|L x B| = |L| |B| sinθ

where θ is the angle between L and B. Since L is in the j direction and B has i and k components, we have:

|L x B| = L * Bz = (3.8 × 10^-3 m) * (1.1 × 10^-4 T) = 4.18 × 10^-8 N

Then, Fx = I * |L x B| = (2.54 × 10^-6 A) * (4.18 × 10^-8 N) = 1.06 × 10^-13 N

To calculate the k component of the magnetic force, we use the same formula and take the k component of the cross product:

|L x B|k = |L| |B| sin(π/2) = |L| |B| = (3.8 × 10^-3 m) * (6.9 × 10^-5 T) = 2.63 × 10^-7 N

Then, Fz = I * |L x B|k = (2.54 × 10^-6 A) * (2.63 × 10^-7 N) = 6.69 × 10^-14 N

The magnitude of the magnetic force is given by,

F = sqrt(Fx^2 + Fz^2) = sqrt((1.06 × 10^-13 N)^2 + (6.69 × 10^-14 N)^2) = 1.26 × 10^-13 N

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