Question

19) A single turn coil of cross-sectional area 7.2cm^2 is in the a magnetic field of flux desity 0.45T. The field which is perpendicular to the coil, is steadily reduced to 0.0T in 5s. Calculate the induced emf.
a) 0:72ÖV
b) 0:52ÖV
c) 0:47ÖV
d) 0:65ÖV
20) What is the self - inductance of an air-core solenoid, 1m long and 0.05m in diameter, if it has 1400 turns?
a) 5.23mH
b) 4.84mH
c) 3.63mH
d) 2.42mH

Answer 1

Answer on Question #61357-Physics-Electromagnetism

19) A single turn coil of cross-sectional area $7.2\mathrm{cm}^2$ is in the a magnetic field of flux density 0.45T. The field which is perpendicular to the coil, is steadily reduced to 0.0T in 5s. Calculate the induced emf.

a) 0:72ÖV

b) 0:52ÖV

c) 0:47ÖV

d) 0:65ÖV

Solution

$$\varepsilon = - \frac {\Delta \Phi}{\Delta t} = - A \frac {\Delta B}{\Delta t} = 7.2 \cdot 10^{-4} \frac {0.45}{5} = 65 \cdot 10^{-6} V.$$

20) What is the self-inductance of an air-core solenoid, 1m long and 0.05m in diameter, if it has 1400 turns?

a) 5.23mH

b) 4.84mH

c) 3.63mH

d) 2.42mH

Solution

The inductance of a coil of wire is given by

$$L = \frac {\mu_ {0} N ^ {2} A}{l}.$$ $$A = \frac {\pi d ^ {2}}{4}.$$ $$L = \frac {\mu_ {0} \pi N ^ {2} d ^ {2}}{4 l} = \frac {4 \pi \cdot 10^{-7} \pi (1400) ^ {2} (0.05) ^ {2}}{4 (1)} = 4.84 \, \mathrm{mH}$$

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