A coil of area $10 {c m}^{2}$ and $10$ turns is in magnetic field directed perpendicular to the plane and changing at a rate of $10^{8}$ $g a u s s / s$ . The resistance of coil is $20 Ω$ . The current in the coil will be

0.5 A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5 \times 10^{- 3} A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.05 A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5 A

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $5 A$
$i = \frac{e}{R}$
$= \frac{N (Δ ϕ / Δ t)}{R} = \frac{N S (Δ B / Δ t)}{R}$
$= \frac{10 (10 \times 10^{- 4}) (10^{4})}{20} = 5 A$

Suggest Corrections

Similar questions

Q. Magnetic field changed at the rate of

0.4 T s^{- 1}

in a square coil of side

4 c m

. Kept perpendicular to the field. If the resistance of the coil is

2 \times 10^{-} 3 Ω

, then induced current in coil is:

A coil of area 10 cm² of 20 turns is placed in uniform magnetic field of 10⁴ gauss. The normal to the plane of the coil makes an angle of 60^o with the magnetic field. The flux in maxwell through the coil is

Q. A coil of 30 turns of wire each of

10 c m^{2}

area is placed with its plane perpendicular to a magnetic field of

0.1 T .

When the coil is suddenly withdrawn from the field, a galvanometer connected in series with the coil indicated that a

10 μ C

charge passes around the circuit. The combined resistance of the coil and galvanometer is:

Q. A coil of area 5

c m^{2}

with

20

turns is kept under the magnetic field of

10^{3}

Gauss. Normal to the plane of coil makes an angle

30^{o}

with the magnetic field. The flux through the coil is

Join BYJU'S Learning Program

A coil of area 10cm2 and 10 turns is in magnetic field directed perpendicular to the plane and changing at a rate of 108 gauss/s. The resistance of coil is 20Ω. The current in the coil will be

The correct option is D 5Ai=eR=N(Δϕ/Δt)R=NS(ΔB/Δt)R=10(10×10−4)(104)20=5A

A coil of area $10 {c m}^{2}$ and $10$ turns is in magnetic field directed perpendicular to the plane and changing at a rate of $10^{8}$ $g a u s s / s$ . The resistance of coil is $20 Ω$ . The current in the coil will be

The correct option is D $5 A$
$i = \frac{e}{R}$
$= \frac{N (Δ ϕ / Δ t)}{R} = \frac{N S (Δ B / Δ t)}{R}$
$= \frac{10 (10 \times 10^{- 4}) (10^{4})}{20} = 5 A$