A 1.0-m long straight wire is held in a horizontal east-west orientation and is then dropped.Supposing that the Earth's magnetic field at this location is 2.0 * 10^-5 T,due north,what will be the induced emf in the wire 2.0 s after it is released? (Hint: The wire is in free-fall,and you can thus calculate its velocity after the 2.0 s.)

Question

Quizplus · Accepted Answer

The induced emf ($emf$) in a wire moving through a magnetic field is given by $emf = Bvl$, where $B$ is the magnetic field strength, $v$ is the velocity of the wire, and $l$ is the length of the wire. After 2.0 s of free fall, the velocity of the wire due to gravity ($g = 9.8 \, m/s^2$) is $v = gt = 9.8 \times 2 = 19.6 \, m/s$. Given $B = 2.0 \times 10^{-5} \, T$ and $l = 1.0 \, m$, the induced emf is $emf = 2.0 \times 10^{-5} \times 19.6 \times 1.0 = 3.92 \times 10^{-4} \, V$, which rounds to $3.9 \times 10^{-4} \, V$, matching choice A.

A 10-M Long Straight Wire Is Held in a Horizontal East-West