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A single-turn circular loop of wire of radius 55 mm lies in a plane perpendicular to a spatially uniform magnetic field. During a 0.10 s time interval, the magnitude of the field increases uniformly from 350 to 450 mT.

Required:
a. Determine the emf induced in the loop (in V). (Enter the magnitude.) V
b. If the magnetic field is directed out of the page, what is the direction of the current induced in the loop?

2 Answers

4 votes

Answer:

(a) 9.5 mV

(b) clockwise

Step-by-step explanation:

Radius, r = 55 mm

Time, t = 0.1 s

Change in magnetic field, B = 450 - 350 = 100 mT =0.1 T

(a) induced emf is given by


e = A (dB)/(dt)


e = A (dB)/(dt)\\\\e=3.14* 0.055*0.055* (0.1)/(0.1)\\\\e= 9.5 * 10^(-3) V = 9.5 mV

(b) According to the Lenz law, the direction of current is clockwise.

User Shaoz
by
8.3k points
7 votes

Answer:

Step-by-step explanation:

Area of the loop = π x ( 55 x 10⁻³ )²

= 9.5 x 10⁻³ m²

Change in Magnetic flux dφ = 450 x 10⁻³ - 350 x 10⁻³ = 150x 10⁻³ Weber.

time dt =.10 s

emf induced = dφ / dt = 150x 10⁻³ Weber / .10 s

= 1.5 V .

b )

Magnetic field is directed outwards and it is increasing so according to Lenz's law , direction of induced current will be clockwise in the loop.

User Kazi
by
8.7k points
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