Question

A coil of 100 turns and area 5 square centimetre is placed in a magnetic field B = 0.2 T. The normal to the plane of the coil makes an angle of 60° with the direction of the magnetic field. The magnetic flux linked with the coil is(a) 5×10⁻³ Wb(b) 5×10⁻⁵ Wb(c) 10⁻² Wb(d) 10⁻⁴ Wb

Answer 1

number of turns in a coil , N = 100

cross section area of coil, A = 5cm² = 5 × 10^-4 m²

magnetic field , B = 0.2 T

a/c to question, The normal to the plane of the coil makes an angle of 60° with the direction of the magnetic field .

so, magnetic flux , $\phi=NBAcos\60^{\circ}$

= 100 × 0.2 × 5 × 10^-4 × cos60°

= 100 × 1/2 × 10^-4

= 5 × 10^-3 Wb

hence, option (a) is correct.