Question

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. Determine the magnitudes of induced emf ε and current I during this time-interval.; (A) ε=1.0mV and I=2mA (B) ε=2.0mV and I=1mA (C) ε=1.2mV and I=2.1mA (D) ε=1.0mV and I=2.1mA

Answer 1

Answer:

1*10^-3 Volt or 1mV and 2.02 × 10-3 Ampere or 2mA .

Explanation:

Since we know that the sides of square loop is = 10 cm = 0.1 m

Also the area of square loop is given as A = 0.01 m^2,

And the resistance of the square loop given is 0.5 Ω .

The magnetic field, B = 0.1 T.

We get that the square loop is placed vertically in the east-west plane and with magnetic field in north-east direction. Hence, the angle between area vector with the magnetic field will be θ = 45.

So, the flux passing through the loop will be:-

Ф = AB cosθ = (0.01)(0.1)(cos45) = 7.07 × 10-4 Wb .

We know from the question that the magnetic field is decreased to zero in the time t = 0.7 s.

Hence, the induced emf will be, e = -dФ/dt = -(7.07 × 10-4)/0.7

So, on solving the value of  e = -1.01 × 10-3 V or 1mV.

Since we consider only magnitude hence the induced emf 1.01 × 10-3 V .

So, the current in the loop is = (1.01 × 10-3)/0.5 = 2.02 × 10-3 A.

Answer 2

Answer:

Explanation: THANKYOU...

I LOVE TO HELP.