Ncert Exercises & Solutions

5.20 A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 45° with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east.

(a) Determine the horizontal component of the earth’s magnetic field at the location.

(b) The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of 90° in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.

Number of turns in the circular coil, N= 30, Radius of the circular coil,
r= 12 cm = 0.12 m
Current in the coil, I =0.35 A, Angle of dip,

δ

= 45°
(a) The magnetic field due to current 1, at a distance r, is given as:

B = \frac{μ_{0} 2 π N I}{4 π r}

W h e r e, μ_{0} = P e r m e a b i l i t y o f f r e e s p a c e = 4 π \times 10^{- 7} T m A^{- 1}

\begin{array}{l} ∴ B = \frac{4 π \times 10^{- 7} \times 2 π \times 30 \times 0.35}{4 π \times 0.12} \\ = 5.49 \times 10^{- 5} T \end{array}

The compass needle points from west to east. hence, the horizontal component of earth's magnetic field is given as:

B_{H} = B \sin δ = 5.49 \times 10^{- 5} \sin 45 ° = 3.88 \times 10^{- 5} T = 0.388 G

(b) When the current in the coil is reversed and the coil is rotated about its vertical axis by an angle of $90 °$ , the needle will reverse its original direction. In this case, the needle will point from East to West.

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