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A straight wire segment of length 0.92 m carries a current of 20.4 A and is immersed in a uniform external magnetic field of magnitude 0.19 T.

(a) What is the magnitude of the maximum possible magnetic force on the wire segment?

User Walrus
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1 Answer

4 votes

Answer:

The magnitude of the maximum possible force is 3.57 newtons.

Step-by-step explanation:

The magnetic force on a wire is:


\overrightarrow{F}=I(\overrightarrow{L}*\overrightarrow{B})

with I the current, L the length of the wire and B the magnitude magnetic field. So, the magnitude of the force is:


F=ILB\sin\theta (1)

with
\theta the angle between the wire and the magnetic field, the maximum possible value is when the wire and magnetic field are perpendicular to each other
\theta= 90Celsius\,degree then
\sin(90)=1 so by (1):


F_(max)=ILB=(20.4A)(0.92m)(0.19T)


F_(max)=3.57N

User Bjorn Behrendt
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