21. An elastic wire forms a circular loop with radius 6.2 cm, placed in a unifor

Question

21. An elastic wire forms a circular loop with radius 6.2 cm, placed in a uniform magnetic field. The field is perpendicular to the plane of the loop, and B = 2.5 T. The loop expands at a speed of 5.0 cm/s. When the loop's radius becomes 10 cm, the induced er A) 0 B) 0.024 V C) 0.039 V D) 0.049 V E) 0.079 v 22. An emf of 3.0 V is connected to a loop circuit. The loop has a radius R = 1.2 m. A uniform magnetic field exists inside this loop. B-0.50-0.60t, where t is in second and B is in T. T he direction of the net current in the circuit is 3.0V A) clockwise B) counter-clockwise C) dependent on the net resistance D) different at different time t E) unable to be determined xXxx 23. A coil has 300 turns, and total inductance of 18 mH. A current of 5.0 mA flows through the coil, the magnetic flux through the coil is A) 1.0×10-9 wb B) 3.0x10" Wb D) 9.0x10- Wb E) 2.7x10 Wb C) 3.0x10 Wb

Explanation / Answer

21)

Area of the loop, A = pi*r^2

So, magnetic flux through it, Q = B*A = B*pi*r^2

So, induced emf, E = dQ/dt = d/dt (B*pi*r^2)

= B*pi*(2*r)*dr/dt

= 2*B*pi*r*dr/dt

So, when the radius is r = 10 cm = 0.1 m,

E = 2*2.5*pi*0.1*0.05

= 0.079 V

22)

B = 0.5 - 0.6*t

So, flux, Q = B*A = B*(pi*r^2)

So, induced emf = dQ/dt = pi*r^2*dB/dt

= pi*1.2^2*(-0.6)

= -2.71 V

As the sign indicates, the induced emf will add with the external emf of 3 V

So, net current will be clockwise.

23)

Magnetic flux, Q = N*L*I

N = 300,

L = 18 mH = 0.018 H

I = 5 mA

So, Q = 300*0.018*0.005

= 0.27 Wb = 2.7*10^-1 Wb

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