FIGURE 23-8
-A constant magnetic field is applied perpendicular to the plane of a square, single-turn coil with each side equal to 15 cm. If the area of the coil decreases at a rate of 0.20 m²/s, an emf of 20 mV is induced in that coil. What is the strength of the magnetic field applied to the coil?

Question

FIGURE 23-8 FIGURE 23-8 -A constant magnetic field is applied perpendicular to the plane of a square, single-turn coil with each side equal to 15 cm. If the area of the coil decreases at a rate of 0.20 m<sup>2</sup>/s, an emf of 20 mV is induced in that coil. What is the strength of the magnetic field applied to the coil? A) 0.10 T B) 0.20 T C) 0.30 T D) 0.40 T E) 0.50 T

FIGURE 23-8 -A constant magnetic field is applied perpendicular to the plane of a square, single-turn coil with each side equal to 15 cm. If the area of the coil decreases at a rate of 0.20 m<sup>2</sup>/s, an emf of 20 mV is induced in that coil. What is the strength of the magnetic field applied to the coil? A) 0.10 T B) 0.20 T C) 0.30 T D) 0.40 T E) 0.50 T

-A constant magnetic field is applied perpendicular to the plane of a square, single-turn coil with each side equal to 15 cm. If the area of the coil decreases at a rate of 0.20 m²/s, an emf of 20 mV is induced in that coil. What is the strength of the magnetic field applied to the coil?

Quizplus · Accepted Answer

Explanation:The emf induced in a coil is given by the equation:emf = -d(BA)/dtwhere B is the magnetic field strength, A is the area of the coil, and t is time.In this problem, the area of the coil is decreasing at a rate of 0.20 m^2/s, and the emf induced in the coil is 20 mV = 0.020 V. The magnetic field strength is constant, so we can rewrite the equation as:emf = -B(dA/dt)Substituting the given values into the equation, we get:0.020 V = -B(0.20 m^2/s)Solving for B, we get:B = 0.10 TTherefore, the strength of the magnetic field applied to the coil is 0.10 T.

FIGURE 23-8 -A Constant Magnetic Field Is Applied Perpendicular to the Plane