Explanation:
Given that,
Voltage of household line, V = 110 V
Power of the hairdryer, P = 1650 W
During use, the current is about 1.95 cm from the user's hand.
(a) Power is given by :
[tex]P=V\times I\\\\I=\dfrac{P}{V}\\\\I=\dfrac{1650\ W}{110\ V}\\\\I=15\ A[/tex]
(b) Again the power is given by :
[tex]P=\dfrac{V^2}{R}[/tex]
R is resistance of the dryer
[tex]R=\dfrac{V^2}{P}\\\\R=\dfrac{(110)^2}{1650}\\\\R=7.34\ \Omega[/tex]
(c) The magnetic field produced by the dryer at the user's hand is given by :
[tex]B=\dfrac{\mu_o I}{2\pi r}\\\\B=\dfrac{4\pi \times 10^{-7}\times 15}{2\pi \times 1.95\times 10^{-2}}\\\\B=1.53\times 10^{-4}\ T[/tex]
Hence, this is the required solution.
