A uniform magnetic field of 2 T exists in a cylindrical region of radius 10 cm, itsdirection parallel to the axis along east to west. A wire carrying current 6 A in the northto south direction passes through the region. What is the magnitude of the force onthe wire if

(a) the wire intersects the axis

(b) the wire in the N-S direction is lowered from the axis by a distance of 6 cm?

Correct option is B

​$$\textbf{Given:} \\\bullet \, \text{Magnetic field strength: } B = 2 \, \text{T} \\\bullet \, \text{Current: } I = 6 \, \text{A} \\\bullet \, \text{Length of the wire: } L = 20 \, \text{cm} = 0.2 \, \text{m} \\\bullet \, \text{The direction of the magnetic field is parallel to the axis of the cylinder, from east to west.} \\\bullet \, \text{The wire is carrying current from north to south.} \\\textbf{Concept:} \\\text{The magnetic force on the wire is given by:} \\F = BIL \sin \theta \\\text{Where:} \\\bullet \, B \text{ is the magnetic field strength,} \\\bullet \, I \text{ is the current,} \\\bullet \, L \text{ is the length of the wire in the magnetic field,} \\\bullet \, \theta \text{ is the angle between the direction of the magnetic field and the current.} \\\textbf{Case a) When the wire intersects the axis (i.e., the angle between the magnetic field and the current is 90°):} \\\text{In this case, } \theta = 90^\circ, \, \text{so } \sin 90^\circ = 1. \\\text{Substituting the values:} \\F = 2 \times 6 \times 0.2 \times 1 = 2.4 \, \text{N} \\\text{Thus, the force when the wire intersects the axis is:} \\F = 2.4 \, \text{N}$$

Case b) When the wire is lowered from the axis by a distance of 6 cm:

​​$$\text{In this case, we first calculate the new length of the wire in the magnetic field.} \\\text{The radius of the cylindrical region is } 10 \, \text{cm}, \text{ and the wire is lowered by } 6 \, \text{cm. The length of the wire in contact with the magnetic field will change based on the position of the wire.} \\\text{Let the horizontal distance from the axis to the wire be } x, \text{ and the total length of the wire in contact with the magnetic field will be } 2x. \\\text{From the geometry, we can calculate:} \\x = \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8 \, \text{cm} \\\text{Thus, the total length of the wire in contact with the magnetic field is:} \\L = 2x = 2 \times 8 = 16 \, \text{cm} = 0.16 \, \text{m} \\\text{Now, using the formula for the magnetic force:} \\F = BIL \sin \theta \\\text{Since the wire is still perpendicular to the magnetic field, } \theta = 90^\circ, \text{ so } \sin 90^\circ = 1. \\\text{Substituting the values:} \\F = 2 \times 6 \times 0.16 \times 1 = 1.92 \, \text{N} \\\text{Thus, the force when the wire is lowered is:} \\F = 1.92 \, \text{N}$$​​