You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Physics

Magnetic Field Due to a Straight Current Carrying Conductor

A closed coil...

Question

A closed coil consists to $500$ turns has area $4 {c m}^{2}$ and a resistance of $50$ $Ω$ . The coil is kept with its plane perpendicular to a uniform magnetic field of $0.2 W b / m^{2}$ . The amount of charge flowing through the coil if it is rotated through $180^{o}$ is $1.6 \times 10^{- x} C$ , x is

Open in App

Solution

$Δ ϕ = 2 (N B S)$
$Δ q = \frac{Δ ϕ}{R} = \frac{2 N B S}{R}$ $= \frac{2 \times 500 \times 0.2 \times 4 \times 10^{- 4}}{50} = 1.6 \times 10^{- 3} C$

Suggest Corrections

Similar questions

Q. A closed coil consists of 500 turns on a rectangular frame of area 4.0

c m^{2}

and has a resistance of 50 ohms. The coil is kept with its plane perpendicular to a uniform magnetic field of 0.2 wb/

m^{2}

, the amount of charge flowing through the coil if it is turned over. (rotated through

180^{0}

) :-

Q. A closed coil consists of

500

turns on a circular frame of area

4 {cm}^{2}

and has a resistance of

50 Ω

. The coil is kept with plane perpendicular to a uniform magnetic field of

0.2 Wb/m

. The amount of charge flowing through the coil if it is turned over (rotated through

180^{\circ}

) will be

Q. A flat coil,

C

n

turn, area

A

and resistance

R

, is placed in a uniform magnetic field of magnitude

B

. The plane of the coil is initially perpendicular to

B

. The coil is rotated by an angle

θ

about a diameter and charge of amount

Q

flows through it. If the coil rotates with a constant angular velocity

ω

, the emf induced in it:

Q. A flat circular coil of

n

turns, area

A

and resistance

R

is placed in a uniform magnetic field

B

. The plane of coil is initially perpendicular to

B

. When the coil is rotated through an angle of

180^{o}

about one of its diameter, a charge

Q_{1}

flows through the coil. When the same coil after being brought to its initial position, is rotated through an angle of

360^{o}

about the same axis a charge

Q_{2}

flows through it. Then

Q_{2} / Q_{1}

Q. A coil of area of cross section

0.5 m^{2}

with

10

turns is in a plane which is perpendicular to a uniform magnetic field of

0.2 W b / m^{2}

. The flux through the coil is:

Join BYJU'S Learning Program

A closed coil consists to 500 turns has area 4cm2 and a resistance of 50 Ω. The coil is kept with its plane perpendicular to a uniform magnetic field of 0.2Wb/m2. The amount of charge flowing through the coil if it is rotated through 180o is 1.6×10−xC, x is

Δϕ=2(NBS)Δq=ΔϕR=2NBSR=2×500×0.2×4×10−450=1.6×10−3C

$Δ ϕ = 2 (N B S)$
$Δ q = \frac{Δ ϕ}{R} = \frac{2 N B S}{R}$ $= \frac{2 \times 500 \times 0.2 \times 4 \times 10^{- 4}}{50} = 1.6 \times 10^{- 3} C$