Ncert Exercises & Solutions

4.23 A uniform magnetic field of 1.5 T exists in a cylindrical region of radius 10.0 cm, its direction parallel to its 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,

Magnetic field strength,

$B = 1.5 T$

Radius of the cylindrical region,

$r = 10 cm = 0.1 m$

Current in the wire passing through the cylindrical region,

$I = 7 A$

(a) If the wire intersects the axis, then the length of the wire is the diameter of the cylindrical region. Thus,

$l = 2 r = 0.2 m$

Angle between magnetic field and current,

$θ = 90 °$

Magnetic force acting on the wire is given by the relation,

$F = BIl \sin θ = 1.5 \times 7 \times 0.2 \times \sin 90 ° = 2.1 N$

Hence, a force of 2.1 N acts on the wire in a vertically downward direction.

(b) New length of the wire after turning it to the Northeast-Northwest direction can be given as:

$l_{1} = \frac{l}{\sin θ}$

Angle between magnetic field and current,

$θ = 45 °$

Force on the wire,

$F = {BIl}_{1} \sin θ$

$= BIl$

$= 1.5 \times 7 \times 0.2$

$= 2.1 N$

Hence, a force of 2.1 N acts vertically downward on the wire. This is independent of angle θ because lsinθ is fixed.

(c) The wire is lowered from the axis by distance,

$d = 6.0 cm$

Let l₂ be the new length of the wire.

$∴ {(\frac{l_{2}}{2})}^{2} = 4 (d + r)$

$= 4 (10 + 6) = 4 (16)$

$∴ l_{2} = 8 \times 2 = 16 cm = 0.16 m$

Magnetic force exerted on the wire,

$F_{2} = {BIl}_{2}$

$= 1.5 \times 7 \times 0.16$

$= 1.68 N$

Hence, a force of 1.68 N acts in a vertically downward direction on the wire.

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