The magnetic field B at the center of a circular coil of wire carrying a current

Question

The magnetic field B at the center of a circular coil of wire carrying a current I (as in fig. 20-9) is where N is the number of loops in the coild and r is its radius. suppose that an electromagnet uses a coild 1.2 m in diameter made from square copper wire 1.6 mm on a side. The power supply produces 120 V at a maximum power output of 4.0 kW.

a) How many turns are needed to run the power supply at maximum power?

b) What is the magnetic field strength at the center of the coil?

c) if you use a greater number of turns and this same power supply (so the voltage remains at 120V), will a greater magnetic field strength result? explain

B=moNI / 2r

Explanation / Answer

(a) We find the resistance of the coil from P = V2/R;

4.0x103 W = (120 V)2/R, which gives R = 3.60 W.

If the coil is tightly wound, each turn will have a length of pD. We find the number of turns

from the length of wire required to give this resistance:

R = rL/A = rNpD/w2;

3.60 W = (1.65x10

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