A square loop of side 10 cm and resistance $0.5 Ω$ is placed vertically in the east-west plane. A uniform magnetic field of 0.10 T is set up across the plane in the north-east direction. The magnetic field is decreased to zero in 0.70 s at a steady rate. The magnitudes of induced emf and current during this time-interval is then

1 mV , 2mA

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1 mV , 1 ma

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2mV , 2mA

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2mV , 1 mA

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Solution

The correct option is A 1 mV , 2mA
$ϕ = B A cos θ$

$= \frac{0.1 \times 10^{- 2}}{\sqrt{3}}$

$ϕ_{min} = 0$

$\in= \frac{ϕ B}{t} = \frac{10^{- 3}}{\sqrt{2} \times 0.7} = 1.0 \times 10^{- 3} = 1 m V$

$I = \frac{\in}{R} = \frac{10^{- 3}}{0.5} = 2 m A$

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Similar questions

Q. A square loop of side

10 c m

and resistance

0.5 Ω

is placed vertically in the east-west plane. A uniform magnetic field of

0.10 T

is set up across the plane in the north-east diraction. The magetic field is decreased to zero in

0.70 s

at a steady rate. the magnitude of current in this time-interval is.

A square loop of side 10cm and R=0.70 ohm is placed vertically in the east-west plane.A uniform magnetic field of 0.10T is set up across the plane in the north-east direction.The magnetic field is decreased to zero in 0.70s at a steady rate.Determine the magnitudes of induced emf and current during this time interval.

Actually my main doubt is that, I know that the value of theta between B and A will be 45° but I don't know that how the value came to be 45° here.

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A square loop of side 4 cm is lying on a horizontal table. A uniform magnetic field of 0.5 T is directed downwards at an angle of $60^{o}$ to the vertical as shown in Fig. 25.3. If the field increases from zero to its final value in 0.2 s, the emf induced in the loop will be

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The correct option is A 1 mV , 2mAϕ=BAcosθ­­­­­­­­­­ =0.1×10−2√3 ϕmin=0 ∈=ϕBt=10−3√2×0.7=1.0×10−3=1mV I=∈R=10−30.5=2mA

The correct option is A 1 mV , 2mA
$ϕ = B A cos θ$

$= \frac{0.1 \times 10^{- 2}}{\sqrt{3}}$

$ϕ_{min} = 0$

$\in= \frac{ϕ B}{t} = \frac{10^{- 3}}{\sqrt{2} \times 0.7} = 1.0 \times 10^{- 3} = 1 m V$

$I = \frac{\in}{R} = \frac{10^{- 3}}{0.5} = 2 m A$