select the correct answer.\nif the emf produced in a wire is 0.88 volts and the wire moves perpendicular to…

select the correct answer.\nif the emf produced in a wire is 0.88 volts and the wire moves perpendicular to a magnetic field of strength 0.075 newtons/amp - meter at a speed of 4.20 meters/second, what is the length of the wire in the magnetic field?\na. 0.27 meters\nb. 2.8 meters\nc. 0.020 meters\nd. 3.7 meters

select the correct answer.\nif the emf produced in a wire is 0.88 volts and the wire moves perpendicular to a magnetic field of strength 0.075 newtons/amp - meter at a speed of 4.20 meters/second, what is the length of the wire in the magnetic field?\na. 0.27 meters\nb. 2.8 meters\nc. 0.020 meters\nd. 3.7 meters

Answer

Explanation:

Step1: Recall the EMF formula

The formula for motional - EMF is $\epsilon = Blv$, where $\epsilon$ is the electromotive force (EMF), $B$ is the magnetic - field strength, $l$ is the length of the wire in the magnetic field, and $v$ is the speed of the wire. We need to solve for $l$.

Step2: Rearrange the formula for $l$

From $\epsilon = Blv$, we can solve for $l$ by dividing both sides of the equation by $Bv$. So, $l=\frac{\epsilon}{Bv}$.

Step3: Substitute the given values

We are given that $\epsilon = 0.88\ V$, $B = 0.075\ N/(A\cdot m)$, and $v = 4.20\ m/s$. Substituting these values into the formula $l=\frac{\epsilon}{Bv}$, we get $l=\frac{0.88}{0.075\times4.20}$. First, calculate $0.075\times4.20 = 0.315$. Then, $l=\frac{0.88}{0.315}\approx2.8\ m$.

Answer:

B. 2.8 meters