2 4 r 1 dq de r 2 4 0 r permeability

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Unformatted text preview: − θ ) = − cot (θ ) = R R π – θ r θ ds=dy y Physics 2B - Summer Session 2 - C. Palmer Current: I to the right August 26, 2013 42 z   Using some geometry we can substitute 1 sin (θ ) = r R out r. x y µ0 I sin θ dy µ0 I (sin θ ) dy ˆ ˆ dB = x= x 2 2 4π r 4π R 3 y cot (π − θ ) = − cot (θ ) = R r π – θ θ ds=dy Current: I to the right y r R Physics 2B - Summer Session 2 - C. Palmer August 26, 2013 43 z   Now we need to substitute out dy. 3 0 2 µ I (sin θ ) dy ˆ dB = x 4π R x y y dy Rdθ 2 = − cot (θ ) ⇒ = ( cscθ ) dθ ⇒ dy = 2 R R (sinθ ) r R π – θ r θ ds=dy y Physics 2B - Summer Session 2 - C. Palmer Current: I to the right August 26, 2013 44 z   Final we can do the integral. 3 0 2 µ I (sin θ ) ⎛ Rdθ ⎞ ˆ x dB = 2⎟ x ⎜ 4π R ⎝ (sin θ ) ⎠ µ 0 I θ =π ˆ B= x ∫ sin θ dθ 4π R θ =0 µ0 I r ˆ = x R 2π R r π – θ θ ds=dy y Physics 2B - Summer Session 2 - C. Palmer y Current: I to the right August 26, 2013 45   The RHR for cross products always works, but this one is faster for B-fields from wires.   Right Hand (Grip) Rule   Use your right hand to grip the wire   Your thumb is in the direction of the of the current   Your fingers point in the direction of the field Physics 2B - Summer Session 2 - C. Palmer August 26, 2013 46...
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This note was uploaded on 09/10/2013 for the course PHYS 2B 2b taught by Professor Hirsch during the Summer '10 term at UCSD.

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