A circular ring of diameter 20 cm has a resistance of 0.01 Ω. The charge that will flow through the ring if it is turned from a position perpendicular to a uniform magnetic field of 2.0 T to a position parallel to the field is about

Question

A circular ring of diameter 20 cm has a resistance of 0.01 Ω. The charge that will flow through the ring if it is turned from a position perpendicular to a uniform magnetic field of 2.0 T to a position parallel to the field is about (A) 63 C (B) 0.63 C (C) 6.3 C (D) 0.063 C

Tardigrade Admin · Accepted Answer

As q =(-dφ/R)=(BA(cos 0°-cos 90°)/R) =(Bπ r2(1+0)/R)=(Bπ r2/R) =(2×3.14×(10-1)2/0.01) =6.28 C ≃ 6.3 C

Q. A circular ring of diameter $20 c m$ has a resistance of $0.01 Ω$ . The charge that will flow through the ring if it is turned from a position perpendicular to a uniform magnetic field of $2.0 T$ to a position parallel to the field is about

63 C

0.63 C

6.3 C

0.063 C

As q $= \frac{- d ϕ}{R} = \frac{B A ( cos 0 ^{\circ} - cos 9 0 ^{\circ} )}{R}$
$= \frac{B π r ^{2} ( 1 + 0 )}{R} = \frac{B π r ^{2}}{R}$
$= \frac{2 \times 3.14 \times ( 1 0 ^{- 1} ) ^{2}}{0.01}$
$= 6.28 C ≃ 6.3 C$

Q. A circular ring of diameter 20cm has a resistance of 0.01Ω. The charge that will flow through the ring if it is turned from a position perpendicular to a uniform magnetic field of 2.0T to a position parallel to the field is about