A coil of wire containing N turns is in an external magnetic field that is perpe

Question

A coil of wire containing N turns is in an external magnetic field that is perpendicular to the plane of the coil and it steadily changing. Under these circumstances, an emf epsilon is induced in the coil. If the rate of change of the magnetic field is doubled and the number of turns in the coil is halved (but nothing else changes), what will be the induced emf in the coil? 2epsilon epsilon/2 4 epsilon epsilon/4 epsilon A flat circular loop of radius 0.20 m is rotating in a uniform magnetic field of 0.50 T. Find the magnetic flux through the Ioop when the plane of the loop and the magnetic field vector are parallel. 2.5 Times 10^-3 T middot m^2 1.5 Times 10^-3 T middot m^2 0.5 Times 10^-3 T middot m^2 0T middot m^2 A magnetic field exerts a force on a charged particle: if the particle is at rest never if the particle is moving along the field lines if the panicle is moving across the field lines always Faraday's law states that an induced emf is proportional to: the rate of change of the electric flux the rate of change of the magnetic flux the rate of change of the electric field the rate of change of the magnetic field zero Two parallel wires carrying equal currents of 10 A attract each other with a force of 1 mN. If one current is halved and the other is doubled, the force of attraction will be: 1 mN 4 mN 0.5 mN 0.25 mN 2 mN A solenoid is 5.0 cm long and has a radius of 0.25 cm. It is wrapped with 500 turns of wire carrying a current of 1.0 A. The magnetic Held at the center of the solenoid is (mu_0 = 4 pi Times 10^-7 T* m/A): 3.3 middot 10^- T 2.3 middot 10^-3 T 1.3 middot 10^-2 T 2.3 middot 10^-2 T 3.3 middot 10^-2 T An electron is moving south in a region where the magnetic field is north. The magnetic force exerted on the electron is: up down cast west

Explanation / Answer

induced emf E = N*(dB/dt)*A

E1 = N1*(dB/dt)1*A1

N1 = N/2

(dB/dt)1 = 2*(dB/dt)

A1 = A

E1 = N/2*2*(dB/dt)*A

E1 = E

option (E)

++++++++++++++++++++

magnetic flux = B.A*costheta

theta = 90

flux = 0

__________________

magnetic force Fb = q*(vxB)

force acts when the charge is in moving across the field

++++++

(8)

force between the two wires

F = uo*I1*I2*L/(2*pi*r)

I1 = I2 = I

F = uo*I^2*L/(2*pi*r)

if I1 = I/2

I2 = 2I

F' = uo*I/2*2I*L/(2*pi*r)

F' = uo*I^2*L/(2*pi*r)

F' = F = 2 mN

option E

++++++++++++++

(9)

Bc = uo*N/L*I

Bc = 4*pi*10^-7*500/0.05*1

B = 1.3*10^-2 T

++++++++++++

10)

Fb = q*(v x B)

Fb = q*v*B*sintheta

theta = 180

Fb = 0

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