$A$ uniform magnetic field of $1.5\; T$ exists in a cylindrical region of radius $10.0\; cm$,its direction parallel to the axis along east to west. $A$ wire carrying current of $7.0\; A$ in the north to south direction passes through this region. What is the magnitude and direction of the force on the wire if,
$(a)$ the wire intersects the axis,
$(b)$ the wire is turned from $N-S$ to northeast-northwest direction,
$(c)$ the wire in the $N-S$ direction is lowered from the axis by a distance of $6.0 \;cm?$

(A) Magnetic field strength,$B = 1.5 \; T$
Radius of the cylindrical region,$r = 10 \; cm = 0.1 \; m$
Current in the wire,$I = 7.0 \; A$
$(a)$ If the wire intersects the axis,the length of the wire inside the field is the diameter,$l = 2r = 0.2 \; m$. The angle between the magnetic field (East-West) and current (North-South) is $\theta = 90^{\circ}$.
Force $F = B I l \sin \theta = 1.5 \times 7.0 \times 0.2 \times \sin 90^{\circ} = 2.1 \; N$. By Fleming's Left-Hand Rule,the direction is vertically downward.
$(b)$ When the wire is turned,the length $l'$ inside the field becomes $l' = l / \sin \theta$,where $\theta = 45^{\circ}$.
Force $F = B I l' \sin \theta = B I (l / \sin \theta) \sin \theta = B I l = 1.5 \times 7.0 \times 0.2 = 2.1 \; N$. The force remains $2.1 \; N$ vertically downward.
$(c)$ If the wire is lowered by $d = 6.0 \; cm$,the length of the chord is $l'' = 2 \sqrt{r^2 - d^2} = 2 \sqrt{10^2 - 6^2} = 2 \sqrt{64} = 16 \; cm = 0.16 \; m$.
Force $F = B I l'' = 1.5 \times 7.0 \times 0.16 = 1.68 \; N$. The direction is vertically downward.