1. Consider two current carrying wires arranged as shown in figure. One wire car
ID: 2007423 • Letter: 1
Question
1. Consider two current carrying wires arranged as shown in figure. One wire carries 4.2A to the right, and the other carries 2.7A upwards. Determine the magnetic field at points A and B.
2. A circular loop moves with constant velocity through regions where uniform magnetic fields of the same magnitude are directed either into or out of the plane of the page. Outside of these two regions, the field is zero. (THe loop is moving from position #1 to #2, etc.) At which of the seven indicated loop positions will the induced current in the loop be..
...Clockwise? _____________
...counterclockwise? _______________
...zero? ______________
Explanation / Answer
The current through wires I1 = 4.2A and through another wire I2 = 2.7A Now from the figure The magnetic field at point A is B = B1 + B2 here B1 = 0 I1 / 2d = (4*10^-7)(4.2)/2(9*10^-2) = 9.33*10^-6 T (out of the page) and B2 = 0 I2 / 2d = (4*10^-7)(2.7)/2(9*10^-2) = 6*10^-6 T (out of the page) therefore the net field B = (9.33 + 6)*10^-6 = 15.33*10^-6 T similarly at point B is B = B1 + B2 here B1 = 0 I1 / 2d = (4*10^-7)(4.2)/2(9*10^-2) = 9.33*10^-6 T (out of the page) and B2 = 0 I2 / 2d = (4*10^-7)(2.7)/2(9*10^-2) = 6*10^-6 T (into the page) B = (9.33-6)*10^-6 = 3.33*10^-6 T (2) The induced current is in opposite direction to the change which will dring it so the current in loop is clockwise then the induced current in counterclok wise. At position 1 it is zero. and at position 2 is in counterclockwise direction. At position 3 there is no change in mangetic field so induced emf is zero. Again at positoin four it have induced emf in positve direction and also in opposaite direction so it is zero. ie. in clockwise direction at position 5 there is no change in mangetic field so induced emf is zero. At positon 6 also there is colckwise direction At position 7 there is no field then no incuced emf . (zero) = (4*10^-7)(2.7)/2(9*10^-2) = 6*10^-6 T (out of the page) therefore the net field B = (9.33 + 6)*10^-6 = 15.33*10^-6 T similarly at point B is B = B1 + B2 here B1 = 0 I1 / 2d = (4*10^-7)(4.2)/2(9*10^-2) = 9.33*10^-6 T (out of the page) and B2 = 0 I2 / 2d = (4*10^-7)(2.7)/2(9*10^-2) = 6*10^-6 T (into the page) B = (9.33-6)*10^-6 = 3.33*10^-6 T (2) The induced current is in opposite direction to the change which will dring it so the current in loop is clockwise then the induced current in counterclok wise. At position 1 it is zero. and at position 2 is in counterclockwise direction. At position 3 there is no change in mangetic field so induced emf is zero. Again at positoin four it have induced emf in positve direction and also in opposaite direction so it is zero. ie. in clockwise direction at position 5 there is no change in mangetic field so induced emf is zero. At positon 6 also there is colckwise direction At position 7 there is no field then no incuced emf . (zero) here B1 = 0 I1 / 2d = (4*10^-7)(4.2)/2(9*10^-2) = 9.33*10^-6 T (out of the page) and B2 = 0 I2 / 2d = (4*10^-7)(2.7)/2(9*10^-2) = 6*10^-6 T (into the page) B = (9.33-6)*10^-6 = 3.33*10^-6 T (2) The induced current is in opposite direction to the change which will dring it so the current in loop is clockwise then the induced current in counterclok wise. At position 1 it is zero. and at position 2 is in counterclockwise direction. At position 3 there is no change in mangetic field so induced emf is zero. Again at positoin four it have induced emf in positve direction and also in opposaite direction so it is zero. ie. in clockwise direction at position 5 there is no change in mangetic field so induced emf is zero. At positon 6 also there is colckwise direction At position 7 there is no field then no incuced emf . (zero) = (4*10^-7)(2.7)/2(9*10^-2) = 6*10^-6 T (into the page) B = (9.33-6)*10^-6 = 3.33*10^-6 T (2) The induced current is in opposite direction to the change which will dring it so the current in loop is clockwise then the induced current in counterclok wise. At position 1 it is zero. and at position 2 is in counterclockwise direction. At position 3 there is no change in mangetic field so induced emf is zero. Again at positoin four it have induced emf in positve direction and also in opposaite direction so it is zero. ie. in clockwise direction at position 5 there is no change in mangetic field so induced emf is zero. At positon 6 also there is colckwise direction At position 7 there is no field then no incuced emf . (zero)Related Questions
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