MASSCHUSETTS INSTITUTE OF TECHNOLOGY ESG Physics. Problem Set 8 Solution

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1 MASSCHUSETTS INSTITUTE OF TECHNOLOGY ESG Physics 8.0 with Kai Spring 003 Problem : 30- Problem Set 8 Solution Determine the magnetic field (in terms of I, a and b) at the origin due to the current loop shown below. The magnetic field at point P due to a finite length is = ( cosθ cosθ) (.) 4π r where θ and θare the angles shown above. So, to find the magnetic field at O, we can use the above formula. We let point P to be the origin, and θ be the angle nearer to the x-axis. The problem can be divided into 3 parts:

2 Part I Consider the left part of the wire. When the current is infinitely far away from the x-axis, θ 0, so cosθ = (.) and d cosθ = (.3) d + a Therefore 0 0 I ( cos cos ) I d = θ θ = (.4) 4πa 4πa d + a And is out of page Part II Consider the part of the wire which is parallel to the x-axis. Denote θ to be the angle on the left. We have r = d and a cosθ = (.5) a + d a cosθ = cos( π θ) = cosθ = (.6) a + d Therefore, is into the page a a a = + = 4π d a + d a + d π d a + d (.7) Part III Consider the part of the wire on the right. It is not difficult to observe that = (.8) 3 As a result, the total magnitude of the magnetic field is tot = + + = + 3 (.9) d? a = k k (.0) π a a + d π d a + d ( )? a = d + a d k k (.) πa a + d πd a + d I ( d a ) d d a 0 = + π ad a + d k ˆ (.)

3 Problem : 30-8 In the figure below, both currents are in the negative x direction. (a) Sketch the magnetic field pattern in the yz plane. (b) At what distance d along the z axis is the magnetic field a maximum? Solution (a) Currents are into the paper (b) Combining the contribution of the magnetic field from each wire, the z components cancel each other on the z axis. Therefore = ˆj (.) Denote as the magnitude of the contribution of magnetic field by one of the wire, then z z = sinθ = = (.) π r r π ( a + z ) where r is the distance from the point on the z axis to one of the wire. Therefore = π Iz 0 ( a + z ) (.3)

4 d To find the maximum of, we set 0 dz =, so we get d z = 0 = dz π a + z a + z a z 0 = π ( a + z ) (.4) (.5) which gives z = a (.6)

5 Problem A solenoid.50 cm in diameter and 30.0 cm long has 300 turns and carries.0 A. (a) Calculate the flux through the surface of a disk of radius 5.00 cm that is positioned perpendicular to and centered on the axis of the solenoid. (b) Figure P30.34b shows an enlarged end view of the same solenoid. Calculate the flux through the blue area, which is defined by an annulus that has an inner radius of cm and outer radius of cm. (a) Φ = A = A (3.) where A is the cross sectional area of the solenoid.. Then oni Φ = ( π r ) = 7.40 Wb (3.) (b) 0NI Φ ( ( = A = A= π r r )) (3.3) which gives Φ =.7 Wb (3.4)

6 Problem 4: Consider the hemispherical closed surface in Figure P If the hemisphere is in a uniform magnetic field that makes an angle θ with the vertical, calculate the magnetic flux (a) through the flat surface S and (b) through the hemispherical surface S (a) Φ, flat = A = π R cos( 80 θ) = πrcosθ (4.) (b) The net flux out of the closed surface is zero, therefore Φ curved, = Φ, flat = π R cosθ (4.)

7 Problem 5: 30.6 (a) I π r 0 4 = =.74 0 T (5.) (b) At point C, conductor A produces a field ( T )( ) produces a field of ( T )( ) j, conductor DE j, D produces no field, and AE produces 4 negligible field. The totle field at C is.74 0 T ( ) j (c) 3 F = IL =.5 0 N (5.) (d) F - a= = ( 0.384ms ) i (5.3) m (e) The bar is already so far from AE that it moves through nearly constant magnetic field. The force acting on the bar is constant, and therefore the bar s acceleration is constant (f) v = v + ax (5.4) f i this gives v = 0.999m/s( i ) (5.5) f

8 Problem 6 : A wire is bent into the shape shown below and the magnetic field is measures at P when the current in the wire is I. The same wire is then formed into the shape shown below, and the magnetic field is measured at P when the current is again I. If the total length of wire is the same in each case, what is the ratio of? For the first figure, we can divide the wire into four equal segments with length l. Then, we can apply equation (.) with θ = 90 and θ = 35 for the four segments, so the total magnetic field at P is 4 ( cos cos ) = θ θ 0 4πl = πl + = (7.) π l For the second figure, since the length of wire is the same, we have π R = 4l, so 4l R = (7.) π For a circular loop of wire, the magnitude of magnetic field at the center is θ = (7.3) 4π R Since in this case, θ = π, thus ( π ) π = = (7.4) 4l 6l 4π π Therefore

9 π l 8 = =.5 π π 6l (7.5)

10 Problem 7: 3-0 Example Consider the arrangement shown below. Assume that R= 6.00 Ω, l =.0m, and a uniform.50-t magnetic field is directed into the page. At what speed should the bar be moved to produce a current of A in the resistor? Since we know that ε = lv (9.) so we can express the current by ε lv I = = (9.) R R Substituting the given values, we can find that v =.00 m/s (9.3)