A circular coil of radius 10 ~cm and resistance of 2 is placed with its plane perpendicular to the horizontal component of the earth's magnetic field. It is rotated about its vertical diameter through 180^ in 0.25 ~s . If the magnitude of the induced emf is 3.8 10⁻³ ~V , then the number of turns of the coil is (Horizontal component of earth's magnetic field at the place is 3 10⁻⁵ ~T )

Step-by-step solution

Radius of coil, r =10 ~cm =0.1 ~m Resistance of coil, R=2 Angle of rotation, =180^ Time of rotation, T =0.25 sec Inducided emf, E =3.8 10⁻³ ~V Horizontal Magnetic field, B _ H =3 10⁻⁵ ~T Induced emf is given as, aligned & E=- N ( _f- _i ) t & = N ( _i- _f ) t & = N (B A ^ -B A ) t aligned ^ = Initial angle =0^ Substitute necessary values, 3.8 10⁻³= N [3 10⁻⁵ (0.1)^2+3 10⁻⁵ (0.1)^2 ] 0.25 on rearranging & solving, Number of turns N =504