Question

A long strait wire lies on the x axis and carries a 3.8 A current in the +x direction. The wire is placed in a uniform magnetic field B=(0.22i−0.34j+0.13k)T. The field will exert a force on the wire.

Part A

Calculate the x-component of the force on the wire per cm of length.

F_x = __________N/cm

Part B

Calculate the y-component of the force on the wire per cm of length.

    Fy = __________N/cm

Part C

Calculate the z-component  of the force on the wire per cm of length.

Fz = __________N/cm

Answer 1

We know that the force on the current carrying conductor placed in a magnetic field (F) = I L x B

That is, F = B * I * L * sin(theta)

Part A

F_x = B_x * I * L * sin(theta)

As the current is in x-direction, the angle between x component of field and x axis will be 0

=> sin(theta) = sin 0 = 0

=> F_x / L = 0

x-component of the force on the wire per cm of length = 0

Part B

F_y = B_y * I * L * sin(theta)

As the current is in x-direction, the angle between y component of field and x axis will be 90

=> sin(theta) = sin 90 = 1

=> F_y / L = -0.34 * 3.8 * 1 = -1.292 N/m = -0.01292 N / cm

y-component of the force on the wire per cm of length = -0.01292 N / cm

Part C

F_z = B_z * I * L * sin(theta)

As the current is in x-direction, the angle between z component of field and x axis will be 90

=> sin(theta) = sin 90 = 1

=> F_z / L = 0.13 * 3.8 * 1 = 0.494 N/m = 0.00494 N / cm

z-component of the force on the wire per cm of length = 0.00494 N / cm