# The right hand rule tells us the direction is into

• Homework Help
• 13
• 100% (1) 1 out of 1 people found this document helpful

This preview shows page 8 - 11 out of 13 pages.

The right-hand rule tells us the direction is into the paper. Due to the fact that A is the same distance from both wires, the total magnetic field at A is B A = 2 µ 0 I 0 π r + µ 0 I 0 π r = 3 µ 0 I 0 π r .
vo (jmv2332) – Magnetic Force and Field HW – yeazell – (58010) 9 Now, the field at B due to the upward current is B up,B = 2 µ 0 I 0 2 π (3 r/ 2) = 2 µ 0 I 0 3 π r again into the paper, while the downward current gives B down,B = µ 0 I 0 2 π ( r/ 2) = µ 0 I 0 π r out of the paper. So at B, the net field out of the paper is: B B = B down,B B up,B = µ 0 I 0 π r parenleftbigg 1 2 3 parenrightbigg = µ 0 I 0 3 π r . Comparing their magnitudes, we find B A B B = 3 µ 0 I 0 π r µ 0 I 3 π r = 9 . 020 10.0points Consider a long wire and a rectangular current loop. A B C D I 1 b a I 2 Determine the magnitude and direction of the net magnetic force exerted on the rectan- gular current loop due to the current I 1 in the long straight wire above the loop. B = µ 0 I 1 2 π r , and the direction of the magnetic field is given by the right hand rule; the field curls around the straight wire with the field coming out of the page above the wire and the field going into the page below the wire. We can now find the force on the segment AB ; applying the right hand rule to find the direction of the cross product, dvectors × vector B , we see that the force will be in the up direction. Since the wire along the segment AB is straight and always at a right angle to vector B , the cross product simplifies to B ds . Since the magnitude of the magnetic field is constant along segment AB , it can come out of the integral which simplifies to give us the result, 5. vector F = µ 0 I 1 I 2 a 2 π ℓ ( b a ), up 6. vector F = µ 0 I 1 I 2 2 π ( a + b ), up 7. vector F = µ 0 I 1 I 2 2 π a b , down 8. vector F = µ 0 I 1 I 2 2 π parenleftbigg a a + b parenrightbigg , down Explanation: To compute the net force on the loop, we need to consider the forces on segments AB , BC , CD , and DA . The net force on the loop is the vector sum of the forces on the pieces of the loop. The magnetic force on AB due to the straight wire can be calculated by using vector F AB = I 2 integraldisplay B A dvectors × vector B . In order to use this, we need to know the magnitude and direction of the magnetic field at each point on the wire loop. We can apply the Biot-Savart Law. The result of this is that the magnitude of the magnetic field due to the straight wire is
vo (jmv2332) – Magnetic Force and Field HW – yeazell – (58010) 10 Following the same argument, we see that the force on the segment CD is F CD = I 2 bracketleftbigg µ 0 I 1 2 π ( a + b ) bracketrightbigg , and its direction is down. This is because the direction of the current is now in in the opposite direction along segment CD !