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? - Zigya
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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,    l = 0.20 m. Also, θ = 90°
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 by Fleming's left hand rule.
(c)    When the wire is lowered by 6 cm, the length of the wire in the cylindrical magnetic field is 2x.
Now,              
                  straight x squared space equals space 10 squared minus 6 squared
straight x space equals space square root of 64 space equals space 8 space cm
therefore              2 straight x space equals space 16 space cm.
                 straight F subscript 2 space equals space BI l subscript 2 space equals space 1.5 space cross times space 7 space cross times space 0.16 space straight N
space space space space space equals space 1.68 space straight N
The force is directed vertically downwards.

a) Diameter of cylindrical region= 20 cm = 0.20 mClearly,    l = 0
The result is true for any angle between current and direction of straight B with rightwards arrow on top. This is because I sin θ remains constant i.e., 20 cm.

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