Class 12 Physics

CBSE and UP Board

Moving Charges and Magnetism

Chapter-4 Exercise 4.23

Additional Exercise

# A uniform magnetic field of 1.5 T exists in a cylindrical region of radius10.0 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?

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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

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

region. Thus,

l= 2r = 0.2 mAngle between magnetic field and current, 0 = 90°

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

F = BI

lsin 0 = 1.5 x 7 x 0.2 x sin 90° = 2.1 NHence, a force of 2.1 N acts on the wire in a vertically downward direction.

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

l₁ = lsin0Angle between magnetic field and current, 0 = 45°

Force on the wire,

F = BI

l₁sin 0 = Bll= 1.5×7 x 0.2 = 2.1 NHence, a force of 2.1 N acts vertically downward on the wire. This is independent of angle 0 because / sin0 is fixed.

Ans (c).The wire is lowered from the axis by distance, d = 6.0 cm

Let

l₂ be the new length of the wire.Therefore (

l₂/2)² = 4 (10 + 6) = 4 (16)Therefore

l₂ =8 x 2 =16 cm =0.16mMagnetic force exerted on the wire,

F₂= BI

l2= 1.5 x 7 x 0.16 = 1.68 N

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