A circular ring of diameter 20 cm has a resistance of 0.01 Ω. The ring is turned from a position perpendicular to a uniform magnetic field of 2T to a position parallel to the field. The amount of charge flowing through the ring in this process is

Question

A circular ring of diameter 20 cm has a resistance of 0.01 Ω. The ring is turned from a position perpendicular to a uniform magnetic field of 2T to a position parallel to the field. The amount of charge flowing through the ring in this process is (A) π C (B) 2π C (C) 8π C (D) 2C

Tardigrade Admin · Accepted Answer

When the ring is in perpendicular position, magnetic flux linked with it is maximum.
i.e.,

ϕ_{1}

=BA =

B \times π r^{2} = 2 \times π \times 1 0^{- 2} = 2 π \times 1 0^{- 2}

Wb (r = 10 cm = 0.1 m)
When ring is in parallel position, magnetic flux linked with it is zero.
i.e.,

ϕ_{2}

= 0

∴

Change in magnetic flux,

Δ ϕ = ϕ_{1} - ϕ_{2} = 2 π \times 1 0^{- 2}

Wb
Given, R = 0.01

Ω = 1 0^{- 2} Ω

Using, charge,

q = \frac{Δ ϕ}{R}

, we have,

q = \frac{2 π \times 1 0 ^{- 2}}{1 0 ^{- 2}} = 2 π

C

Q. A circular ring of diameter $20 c m$ has a resistance of $0.01$ $Ω$ . The ring is turned from a position perpendicular to a uniform magnetic field of $2 T$ to a position parallel to the field. The amount of charge flowing through the ring in this process is

$π C$

$2 π C$

$8 π C$

2C

Q. A circular ring of diameter 20cm has a resistance of 0.01 Ω. The ring is turned from a position perpendicular to a uniform magnetic field of 2T to a position parallel to the field. The amount of charge flowing through the ring in this process is

πC

2πC

8πC

2C

Q. A circular ring of diameter $20 c m$ has a resistance of $0.01$ $Ω$ . The ring is turned from a position perpendicular to a uniform magnetic field of $2 T$ to a position parallel to the field. The amount of charge flowing through the ring in this process is

$π C$

$2 π C$

$8 π C$