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Question: A boat is moving due east in a region where the earth's magnetic field is 3.6 $\times$ 10$^{-5}$ N/A...

A boat is moving due east in a region where the earth's magnetic field is 3.6 ×\times 10−5^{-5} N/Am due north and horizontal. The boat carries a vertical conducting rod 2 m long. If the speed of the boat is 2.00 m/s, the magnitude of the induced e.m.f. in the rod is

A

1.4 mV

B

0.72 mV

C

0.54 mV

D

0.144 mV

Answer

0.144 mV

Explanation

Solution

The induced emf in a moving conductor is given by:

E=B v L\mathcal{E} = B \, v \, L

Here, the magnetic field B=3.6×10−5 N/AmB = 3.6 \times 10^{-5} \, \text{N/Am}, speed v=2.00 m/sv = 2.00 \, \text{m/s}, and the rod length L=2 mL = 2 \, \text{m}. Plugging in the values:

E=(3.6×10−5)×(2.00)×(2)=1.44×10−4 V=0.144 mV.\mathcal{E} = (3.6 \times 10^{-5}) \times (2.00) \times (2) = 1.44 \times 10^{-4} \, \text{V} = 0.144 \, \text{mV}.