Class 12 Physics - Chapter Moving Charges and Magnetism NCERT Solutions | A uniform magnetic field of 1.5 T exists
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?
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 = 2r = 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 × 7 × 0.2 × 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: :
Angle between magnetic field and current, θ = 45°
Force on the wire,
F = BIl1 sin θ
= BIl
= 1.5 x 7 x 0.2
= 2.1 N
Hence, a force of 2.1 N acts vertically downward on the wire. This is independent of angleθbecause l sinθ is fixed.
(c) The wire is lowered from the axis by distance, d = 6.0 cm
Suppose wire is passing perpendicularly to the axis of cylindrical magnetic field then lowering 6 cm means displacing the wire 6 cm from its initial position towards to end of cross sectional area.
Thus the length of wire in magnetic field will be 16 cm as AB= L =2x =16 cm
Now the force,
F = iLB sin90° as the wire will be perpendicular to the magnetic field.
F= 7 × 0.16 × 1.5 =1.68 N
The direction will be given by right hand curl rule or screw rule i.e. vertically downwards.
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