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Question

Physics Question on Electrical Instruments

Two tangent galvanometer having coils of the same radius are connected in series. A current flowing in them produces deflection of 6060^{\circ} and 4545^{\circ} respectively. The ratio of the number of turns in the coils is

A

4/34/3

B

(3+1)/1(\sqrt{3}+1) / 1

C

3/1\sqrt{3} / 1

D

(3+1)/(31)(\sqrt{3}+1) /(\sqrt{3}-1)

Answer

3/1\sqrt{3} / 1

Explanation

Solution

Tangent galvanometer is an early measuring instrument for small electric currents. It consists of a coil of insulated copper wire wound on a circular non-magnetic frame. Its working is based on the principle of the tangent law of magnetism. When a current is passed through the circular coil, a magnetic field (B)(B) is produced at the center of the coil in a direction perpendicular to the plane of the coil. The TG TG is arranged in such a way that the horizontal component of earth?s magnetic field (Bh)(B_h) is in the direction of the plane of the coil. The magnetic needle is then under the action of two mutually perpendicular fields. If ? is the deflection of the needle, then according to tangent law, B=BhtanθB = B _{ h } \tan \theta where B=μ0nI2aB =\frac{\mu_{0} nI }{2 a } Where n is number of coils, II is current and a is radius of coil. Given radius of both coils are same. The current will be same as both coils are connected in series. B1=μ0n1I2a=Bhtanθ1(i)B _{1}=\frac{\mu_{0} n _{1} I }{2 a }= B _{ h } \tan \theta_{1} \ldots (i) B2=μ0n2I2a=Bhtanθ2(ii)B _{2}=\frac{\mu_{0} n _{2} I }{2 a }= B _{ h } \tan \theta_{2} \ldots(ii) B1B2=tanθ1tanθ2\frac{ B _{1}}{ B _{2}}=\frac{\tan \theta_{1}}{\tan \theta_{2}} n1n2=tanθ1tanθ2\frac{ n _{1}}{ n _{2}}=\frac{\tan \theta_{1}}{\tan \theta_{2}} n1n2=tan60tan45\frac{ n _{1}}{ n _{2}}=\frac{\tan 60}{\tan 45} n1n2=31\frac{ n _{1}}{ n _{2}}=\frac{\sqrt{3}}{1}