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)
Diameter of cylindrical region = 20 cm = 0.20 m
Clearly,
length of conductor, l = 0.20 m.
Also, θ = 90°
Current carried by conductor, I = 7A
Uniform magnetic field, B= 1.5 T 

Force, F = BIl sin θ
            = 1.5 x 7 x 0.20 sin 90° N
            = 2.1 N

Using Fleming's left-hand rule, we find that the force is directed vertically downwards.

(b) If l₁ is the length of the wire in the magnetic field, then,

F₁ = BIl₁ sin 45°

But , l₁ sin 45° = l
∴ F₁ = BIl
       = 1.5 x 7 x 0.20 N
       = 2.1 N 

The force is directed vertically downwards( using Fleming's left hand rule). 

(c) When the wire is lowered by 6 cm from axis, the length of the wire in the cylindrical magnetic field is 2x. 



Now considering the right angled triangle as shown in fig. above,

                   $$\begin{gathered} {\mathrm{x}}^2 = {10}^2-6^2 \\ \mathrm{x} = \sqrt{64} = 8 \mathrm{cm} \end{gathered}$$ 
$\therefore$              $2\mathrm{x} = 16 \mathrm{cm}.$ 

                 $$\begin{gathered} {\mathrm{F}}_2 = \mathrm{BI}{l}_2 \\ = 1.5 \times 7 \times 0.16 \mathrm{N} \\ = 1.68 \mathrm{N} \end{gathered}$$ 

The force is directed vertically downwards.

The result is true for any angle between current and direction of $\overrightarrow{\mathrm{B}}$. This is because I sin θ remains constant i.e., 20 cm.