Question

1.
A vertical circular coil of radius 0.1 m and having
10 turns carries a steady current, when the plane of
the coil is normal to magnetic meridian, a neutral
point is observed at the centre of the coil. If
By = 0.314x10 T, then current in the coil is :​

Answer 1

Explanation:

Given A vertical circular coil of radius 0.1 m and having  10 turns carries a steady current, when the plane of  the coil is normal to magnetic meridian, a neutral  point is observed at the centre of the coil. If  Bh = 0.314 x 10^-4 T, then current in the coil is :

• Now there is a vertical coil with radius r = 0.1 m and n = 10 turns and a steady current I.
• When the plane of coil is normal due to magnetic meridian there is a neutral point B = 0
• Now magnetic field due to current I will be
• B I = μo n I / 2R = B h (so that they cancel each other and get a neutral point)  
•      Or I = 2 Bh R / μo x n
•              = 2 x 0.314 x 10^-4 x 0.1 / 10 x 4 π x 10^-7
•             = 0.314 x 10^-4 x 0.1 / 20 x 3.14 x 10^-7
•           = 0.0314 x 10^-4 / 62.8 x 10^-7
•               = 0.0005 x 10^3
•      Or I = 0.5 A

Reference link will be

https://brainly.in/question/11150049