Moving bar

Electricity and Magnetism Level 2

A metal bar of length 50cm is accelerating at a rate a = 2 t + 3 a=2t+3 , where t t is in seconds. After five seconds, the bar has a velocity of 25m/s. If the bar is travelling perpendicularly through a uniform magnetic field of strength 0.35T, how much voltage is the being generated after twelve seconds?

Notes: Assume there is no friction or air resistance. Give your answer to 3 significant figures.


The answer is 28.9.

Charley Shi by Charley Shi , 19, New Zealand

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1 solution

Since the bar and the magnetic field are perpendicular, the induced voltage is given by the formula : ∣ U ∣ = B ∗ l ∗ v |U| = B * l * v where l is the length of the bar, B the strength of the magnetic field and v the velocity of the relative movement.

a ( t ) = 2 t + 3 a(t) = 2t + 3

so v ( t ) = t 2 + 3 t + C v(t) = t^2 + 3t + C

as v ( t = 5 ) = 25 v(t=5) = 25 , we get C = − 15 C = -15

finally v ( t ) = t 2 + 3 t − 15 v(t) = t^2 + 3t - 15

and U = 0.35 ∗ 0.5 ∗ ( 1 2 2 + 3 ∗ 12 − 15 ) = 28.875 V U = 0.35 * 0.5 * (12^2 + 3*12 -15) = 28.875 V

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