test5_10Sol

# Three locations are marked a b are far away from the

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Unformatted text preview: ugh. Three locations are marked, A, B are far away from the loop, so its contribution can be ignored. Location C is at the centre of the loop. Use the formulae provided to calculate the magnetic ﬁelds at A, B, C in the paper plane. A current of 1.5 A ﬂows through the wire. Be careful with location C , there are two contributions. The ﬁelds are to be speciﬁed by magnitude and direction or by listing the appropriate component with sign! 3 4) [5] The particle in the ﬁgure has a negative charge, and its velocity vector lies in the x − y plane and makes an angle of 75◦ with the y axis. A magnetic ﬁeld is along the +x direction. What is the direction of the magnetic force on the particle? Now you are told that the ﬁeld has a strength of 1.5 T, and that the particle speed is v = 500 m/s. Calculate the magnetic force. 4 FORMULA SHEET ￿ v (tf ) = v (ti ) + ti a(t) dt s(tf ) = s(ti ) + ttif v (t) dt vf = vi + a∆t sf = si + vi ∆t + 1 a∆t2 vf2 = vi2 + 2a∆s 2 ￿ tf g = 9.8 m/s2 f (t) = t f (t) = a = 0 F (t) = f (t) dt = at + C F (t) = anti-derivative = indeﬁnite integral area under the curve f (t) between limits t1 and t2 : F (t2 ) − F (t1 ) x2 + px + q = 0 fac...
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## This note was uploaded on 09/28/2010 for the course MATH 1025 taught by Professor Tammie during the Fall '10 term at York University.

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