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3 years ago

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A metal rod with a length of 22.0cm lies in the xy-plane and makes an angle of 37.9 degrees with the positive x-axis and an angl

e of 52.1degrees with the positive y-axis. The rod is moving in the +x-direction with a speed of 6.80m/s . The rod is in a uniform magnetic field B =(0.170T)i^-(0.240T)j^-(0.0900T )k^.
What is the magnitude of the emf induced in the rod?

1 answer:

Shtirlitz [24]3 years ago

8 0

Answer:

Induced emf in the rod is, $\epsilon=0.08262\ V$

Explanation:

Given that,

Length of the rod, L = 22 cm = 0.22 m

Angle with x axis is 37.9 degrees with the positive x-axis and an angle of 52.1 degrees with the positive y-axis.

$L=0.22(cos37.9\ i)+0.22(sin37.9\ j)$

$L=(0.173\ i+0.135\ j)\ m$

Velocity of the rod, v = 6.8i m/s

Magnetic field, $B=0.170i-0.24j-0.09k$

The formula for the emf induced in the rod is given by :

$\epsilon=(v\times B){\cdot}L$

$\epsilon=(6.8i\times (0.170i-0.24j-0.09k)){\cdot} (0.173\ i+0.135\ j)$

$\epsilon=(0.612j-1.632k){\cdot}(0.173i+0.135j)$

$\epsilon=0.08262\ V$

So, the magnitude of the emf induced in the rod is 0.08262 volts. Hence, this is the required solution.

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Answer:

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<u>Free fall </u>

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