# 1 the induced magnetic field is right to left b

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1. The induced magnetic field is right to left ( = B induced ) and the induced current is zero. 2. The induced magnetic field is left to right ( B induced = ) and the induced current flows from a through R to b ( I -→ ). 3. The induced magnetic field is zero and the induced current flows from a through R to b ( I -→ ). 4. The induced magnetic field is zero and the induced current flows from b through R to a ( ←- I ). 5. The induced magnetic field is left to right ( B induced = ) and the induced current is zero. 6. The induced magnetic field is left to right ( B induced = ) and the induced current flows from b through R to a ( ←- I ). correct 7. The induced magnetic field is zero and the induced current is zero. 8. The induced magnetic field is right to left ( = B induced ) and the induced current flows from a through R to b ( I -→ ). 9. The induced magnetic field is right to left ( = B induced ) and the induced current flows from b through R to a ( ←- I ). Explanation: The induced magnetic field depends on whether the flux is increasing or decreasing. The magnetic flux through the coil is from right to left. When the magnet moves from right to left, the magnetic flux through the coils increases. The induced current in the coil must pro- duce an induced magnetic field from right to left ( = B induced ) to resist any change of magnetic flux in the coil (Lenz’s Law). The helical coil when viewed from the bar magnet winds around the solenoid from ter- minal b clockwise. Since the induced field is right to left ( = B induced ), the induced current flows from b through R to a ( ←- I ); i.e. , counter- clockwise. 008 (part 1 of 2) 10.0 points A cylinder with a(n) 1 . 9 cm radius and a length of 4 . 6 m is tightly wrapped with 3600 turns of wire. The current in the wire is decaying according to I = I 0 e - α t , with I 0 = 1 . 4 A and α = 2 . 5 s - 1 . What is the electric field at a point 9 . 7 cm radially from the axis of the cylinder at t = 0 . 76 s? Correct answer: 9 . 57978 × 10 - 7 V / m. Explanation: Let : r = 1 . 9 cm , = 4 . 6 m , N = 3600 , I 0 = 1 . 4 A , and t 1 = 0 . 76 s . Let us take a circle of radius a around the cylinder. According to Faraday’s law of in-

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karna (pk4534) – HW 08 – li – (59050) 6 duction, contintegraldisplay E · d s = - d Φ B dt . By the symmetry of the situation, we can see that the electric field is constant around the circle and tangent to it. Thus E = - 1 2 π a d Φ B dt . The magnetic field inside the cylinder is es- sentially that of a solenoid, B = μ 0 n I , where n is the number of turns per unit length ( n = N = 3600 4 . 6 m = 782 . 609). Outside the cylinder, the component of the magnetic field parallel to the axis is zero, so the magnetic field outside the solenoid does not contribute to the magnetic flux through the central plane of the cylinder. This magnetic flux is there- fore Φ B = A cyl B cyl = π r 2 μ 0 n I . The time variation of the magnetic flux is then d Φ B dt = π r 2 μ 0 n d I dt . Since I ( t ) = I 0 e - αt , d I dt vextendsingle vextendsingle vextendsingle vextendsingle t = t 1 = - α I 0 e - αt 1 = 0 . 52349 A / s .

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