A straight, vertical wire carries a current of 2.60 Adownward in a region between the poles of a large superconducting electromagnet, where the magnetic field has magnitude B = 0.588 Tand is horizontal. What are the magnitude and direction of the magnetic force on a 1.00 cmsection of the wire that is in this uniform magnetic field, if the magnetic field direction is (a) east; (b) south; (c) 30.0⁰ south of west?

It is a region around charged particle within which a force would be exerted on another charged particle.

Current l = 2.60 A

Magnetic field magnitude B = 0.588 T

Section on wire l = 1.00 cm

(a)

the current direction is downwards from the upper view when magnetic field direction is in east. Angle between length and magnetic field is .by right hand rule force is in south which can be given as,

$F=llB\sin \varphi$

Here I is the current, l is the wire length and B is the magnetic field.

Substitute all the value in the above equation.

$\begin{array}{r}F=\left(2.60\mathrm{A}\right)\left(1.00\times {10}^{-2}\mathrm{m}\right)\left(0.588\mathrm{T}\right)\\ =0.0153\mathrm{N}\\ F=0.0153\mathrm{N}\end{array}$

Thus, the magnetic force F = 0.01533 N in south direction

(b)

In second, the angle between length and magnetic field is$\varphi =90°$ when magnetic field is in south direction.by right hand rule force points to west,

role="math" localid="1668233909743" $F=llB\sin \varphi$

Here I is the current, l is the wire length and B is the magnetic field.

Substitute all the value in the above equation.

$\begin{array}{r}F=\left(2.60\mathrm{A}\right)\left(1.00\times {10}^{-2}\mathrm{m}\right)\left(0.588\mathrm{T}\right)\\ =0.0153\mathrm{N}\\ F=0.0153\mathrm{N}\end{array}$

Thus, the magnetic force in west direction

(c)

In third, magnetic field is toward south and force is perpendicular to magnetic field and length.by right hand rule force direction is north of west by $60°$.

$F=llB\sin \varphi$

Here I is the current, l is the wire length and B is the magnetic field.

Substitute all the value in the above equation.

$\begin{array}{r}F=\left(2.60\mathrm{A}\right)\left(1.00\times {10}^{-2}\mathrm{m}\right)\left(0.588\mathrm{T}\right)\\ =0.0153\mathrm{N}\\ F=0.0153\mathrm{N}\end{array}$

Thus, the magnetic force F = 0.01533 N in North-West direction.