Question

A small current-carrying loop is located at a distance r from a long straight conductor with current I. The magnetic moment of the loop is equal to p_(m). Find the magnitude and direction of the force vector applied to the loop if the vector p_(m) (a) is parallel to the stratight conductor, (b) is oriented along the radius vector r, (c) coincides in direction with the magnetic field produced by the current I the point where the loop is located.

Answer 1

Answer:

Due to the straight conductor, Bφ=μ0I2πr

We use the formula, F→=(pm−→.∇→)B→

(a) The vector pm−→ is parallel to the staright conductor.

F→=pm∂∂ZB→=0,

because netiher the direction nor the magnitude of B→ depends on z

(b) The vector pm−→ is oriented along the radius vector r→

F→=pm∂∂rB→

The direction of B→ at r+dr is parallel to the direction at r. Thus only the φ component of F→ will survive.

Fφ=pm∂∂rμ0I2πr=−μ0Ipm2π2

(c) The vector pm−→ coincides in direction with the magnetic field, produced by the conductor carrying current I

F→=pm∂r∂φμ0I2πeφ→=μ0Ipm2πr2∂eφ→∂φ

So, F→=μ0Ipm2πr2er→ As, ∂eφ→∂φ=−er→