Question

(2) B-field Curve. A circular ring of radius r = 0.10 m and resistance Ro = 2.02 sits in a uniform magnetic field pointing out of the page. The graph shows the strength of that field as a function of time (a negative value indicates that the field is pointing into the page at that moment). (a) At what time (or times) is the largest current induced in the ring? (b) Between t=2s and t = 3 s is the induced current clockwise or counterclockwise? Explain. (c) Between t = 2s and t = 3 s is the induced current increasing, decreasing, or constant in magni- tude? Explain. (d) What is the induced current at t = 4.7s? (e) What is the induced current at t = 4.0 s? (f) Between t = 5 s and t = 6s how is the induced current increasing, decreasing, or constant in magnitude? Explain. B (T) 1.0 2.0 3.0 4.0 5.0 6.0

Answer 1

Induced current ids given by,I=-($\pi$r²/R₀)(dB/dt) ..................(1)

Lenz law:states that the direction of the current induced in a conductor by a changing magnetic field is such that the magnetic field created by the induced current opposes the initial changing magnetic field.

Part a.

Since induced current is proportional to the rate of change og magnetic field.So induced current is proportional to the slope of line.

Slope of line is maximum from t=3 s to t=4.5 s

So From t=3 s to t=4.5 s largest current induced in ring.

Part b.

Because in this interval magnetic field is increasing.According to lenz law induced current id in clockwise direction.

Part c.

From equation 1 induced current is decreasing.

Part d.

Slope of line is zero at t=4.7 s ,So induced current is zero at t=4.7 sec.

Part e.

| 1.0 2.0 3.0 4.0 5.0 6.0

Slope of line at t=4 sec=-(0.2+0.1)/(4.5-3)=-0.2 T/s

So

Induced current at t=4 sec is=,I=-($\pi$r²/R₀)(dB/dt)=-($\pi$$\times$0.10²/2)(-0.2)=3.14$\times$10^-3 A

Part f.

Slope in this interval is positive.From equation 1 induced current is decreasing.