Question #232436

You have a 12V battery connected to a single straight nichrome wire in your heater. If the wire is a millimeter thick and a meter long, and you place a device that measures magnetic field strength at a one-centimeter distance away from the wire, what will be the reading on this device?


1
Expert's answer
2021-09-02T16:46:19-0400

Device is placed very near the wire then wire will behave as a infinite long current carrying wire.

Magnetic field strength due to current carrying wire is given by

B=μ0×i2πdμ0=4π×107B = \mu_0 \times \frac{i}{2 \pi d} \\ \mu_0= 4 \pi \times 10^{-7} \\

d = distance of device from wire = 1 cm = 0.01 m

i = current in wire =ER= \frac{E}{R} (from ohm's law)

E = apply battery voltage = 12 V

R = resistance of wire =ρ×LA= \rho \times \frac{L}{A}

ρ\rho = resistivity of Nichrome = 1.0×106  ohm×m1.0 \times 10^{-6} \; ohm \times m

L = Length of wire = 1 m

A = cross sectional area of wire =πd24= \frac{\pi d^{2}}{4}

d = diameter of wire = thickness of wire = 1 mm = 0.001 m

i=12(1.0×106)×1(0.25×π×0.0012)=9.425  AB=(4×π×107)×9.425(2π×0.01)B=0.0001885  TB=1.885×104  Ti = \frac{12}{\frac{(1.0 \times 10^{-6}) \times 1}{(0.25 \times \pi \times 0.001^2)}} = 9.425 \; A \\ B = \frac{(4 \times \pi \times 10^{-7}) \times 9.425}{(2 \pi \times 0.01)} \\ B = 0.0001885 \; T \\ B = 1.885 \times 10^{-4} \; T


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