Question

1 A magnetic field of 0,5 T passes through an conducting loop of area 10 cm². The field lines are at 70° to the plane of the loop. 1.1 Calculate the magnetic flux linkage. (Hint: First, calculate the angle between the field and the normal to the plane.) 1.2 Calculate the average emf that will be induced across the ends of the coil if it is removed from the field in 0,2 s. (Hint: It is removed from the field so final is zero.) 1.3 Calculate the induced current if the loop is connected to a circuit with a resistance of 2 Q. 2 A solenoid with 400 turns is rotated so that the magnetic flux linked x 10-4 Wb in 0,1 s. 3 with each turn changes from 5 x 10-4 Wb to Calculate the average emf across the ends of the solenoid. A wire loop with an area of 0,02 m² is positioned in a magnetic field in which the flux density changes from 0,1 T to 0,8 T in 0,4 s. The magnetic field is at an angle of 25° to the normal to the plane of the loop. 3.1 Calculate the induced emf. 3.2 Calculate the induced current if the loop is connected to a circuit with a resistance of 5 Q. F i C q 1 3​

Answer 1

1.1 To calculate the magnetic flux linkage, we need to first find the angle between the field and the normal to the plane. Since the field lines are at 70° to the plane of the loop, the angle between the field and the normal is 20°. Therefore, the magnetic flux linkage is:

Magnetic flux linkage = magnetic field * area * cos(angle)

Plugging in the given values, we get:

Magnetic flux linkage = 0.5 T * 10 cm^2 * cos(20°) = 0.087 Wb

1.2 The average emf induced across the ends of the coil is given by:

Average emf = change in magnetic flux linkage / time

Since the coil is removed from the field, the final magnetic flux linkage is zero. Therefore, the change in magnetic flux linkage is equal to the initial magnetic flux linkage, which we calculated in part 1.1. Plugging in the given values, we get:

Average emf = 0.087 Wb / 0.2 s = 0.435 V

1.3 The induced current can be found using Ohm's law:

I = V / R

Plugging in the given values, we get:

I = 0.435 V / 2 Ω = 0.218 A

2. The average emf induced across the ends of the solenoid is given by:

Average emf = change in magnetic flux linkage / time

Since the magnetic flux linked changes from 5 x 10^-4 Wb to 15 x 10^-4 Wb, the change in magnetic flux linkage is:

Change in magnetic flux linkage = 15 x 10^-4 Wb - 5 x 10^-4 Wb = 10 x 10^-4 Wb

Plugging in the given values, we get:

Average emf = 10 x 10^-4 Wb / 0.1 s = 0.1 V

Therefore, the average emf across the ends of the solenoid is 0.1 V.

3.1 The induced emf can be found using Faraday's law:

Induced emf = -N * d(flux) / dt

Since the flux density changes from 0.1 T to 0.8 T, the change in flux density is:

Change in flux density = 0.8 T - 0.1 T = 0.7 T