# When operated on a household 110.0 V line, typical hair dryers draw about 1450 W of power. The current can be modeled as a long, straight wi

Question

When operated on a household 110.0 V line, typical hair dryers draw about 1450 W of power. The current can be modeled as a long, straight wire in the handle. During use, the current is about 2.05 cm from the user’s hand. What is the current in the dryer? What is the resitance of the dryer? What magnetic field does the dryer produce at the users hand?

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1 year 2021-08-29T20:11:45+00:00 1 Answers 38 views 0

1. To solve this problem we will start by calculating the resistance through the relationship between it, the voltage and the power. From there we will apply Ohm’s law and later we will find the current. With these values we will have all the variables for the calculation of the magnetic field. So the resistance would be given by

$$P = \frac{V^2}{R}$$

$$R = \frac{V^2}{P}$$

Replacing with our values

$$R = \frac{(110.0V)^2}{1450W}$$

$$R = 8.345\Omega$$

The expression for Ohm’s law is given below,

$$I = \frac{V}{R}$$

$$I = \frac{110.0V}{8.345\Omega}$$

$$I = 13.18A$$

The magnetic field at any point from the current carrying straight wire depends on the current and the distance between the point and the wire, then

$$B = \frac{\mu_0 I}{2\pi r}$$

Here,

$$\mu$$ = Permeability constant

I = Current

r = Distance between the points

Replacing

$$B = \frac{(4\pi*10^{-7} H\cdot m^{-1})(13.18A)}{2\pi (2.05*10^{-2}m)}$$

$$B = 1226.0*10^{-7}T$$

Therefore the magnetic field is given as $$1226.0*10^{-7}T$$