Physics, asked by Anjuna8993, 11 months ago

A moving coil galvanometer has a coil with 175 turns and area 1 cm²It uses a torsion band of torsion constant 10⁻⁶ N – m/rad. The coil is placed in a magnetic field B parallel to its plane. The coil deflects by 1⁰ for a current of 1 mA. The value of B (in Tesla) is approximately:-
(A) 10⁻³
(B) 10⁻¹
(C) 10⁻⁴
(D) 10⁻²

Answers

Answered by Anonymous
53

{\red{\huge{\underline{\mathbb{Moving  coil  Galvanometer:-}}}}}

The moving engagement was devised by Kelvin and later on modified by D' Arsonaval . this is used for detection and measurement of small electric current.

Theory :

Torque on the coil = niAB

where ,n = number of turns in the coil

A = area

and B = magnetic field induction of radial magnetic field in which the call is suspended.

↪due to the torque , the coil rotates. no restoring torque is developed in the suspension wire.

in equilibrium,

deflecting couple= restoring torque

niAB = C∅

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Given : i = 1 mA

∅ = 1° = π/180 radian

A = 1 cm²

C = 10 {}^{ - 6} N \: m / \: radian

n = 175 turns

C∅= niAB

put the given values

10 {}^{ -6}  \times   \frac{\pi}{180} = 10 {}^{ - 3}   \times 10 {}^{ - 4}  \times 175 \times B

B = 10 {}^{ - 3}  \: tesla

correct option a)

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Answered by anshi60
5

{\red{\huge{\underline{\mathbb{Answer:-}}}}}

Given : i= 1 mA

N = 175 turns

∅ = 1° = π/180 radian

A = 1 cm²

C = 10⁻⁶ N – m/rad.

at equilibrium

C∅= niAB

put the given values

10 {}^{ - 6}  \times  \frac{\pi}{180}  = 10 {}^{ - 3}  \times 10 {}^{ - 4}  \times 175 \times B

B = 2 tesla

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