The wires in the figure carry currents of I1 = 4A and I2 = 3A in the depicted di
ID: 2026027 • Letter: T
Question
The wires in the figure carry currents of I1 = 4A and I2 = 3A in the depicteddirections. The distance between the wires is 2cm. Point P1 is 1cm to the left of I1, point P2 is in the middle point between the currents and point P3 is 1cm to the right of I2.
http://imgur.com/BZycE
a.) Calculate the magnetic field (magnitude and direction) at points, P1, P2 and P3.
P1 :
P2 :
P3 :
b) Calculate the acceleration (magnitude and direction) of an electron
(me = 9.10938188 × 10-31 kg) traveling into the page with a speed of 4 x 105 m/s at point P2 .
c) Calculate the force (magnitude and direction) that the left wire exerts on the right
one.
d) Find a point on the x-axis where the magnetic field vanishes. Give it’s position with
respect to the current I2.
Explanation / Answer
Given: currents in the wires , 1 and 2 is I1 = 4 A I2 = 3A distance between the wires , d = 2 cm = 2 cm (10-2 m /1 cm ) = 0.02 m distance b/w current I1 to point P1 is , r 1 = 1 cm = 1 cm (10-2 m /1 cm ) = 0.01 m distance b/w current I2 to point P3 is , r 3 = 1 cm = 1 cm (10-2 m /1 cm ) = 0.01 m distance b/w I1 or I2 to piont P2 is , r 2 = 2/2 cm = 1 cm (10-2 m /1 cm ) = 0.01 m velocity of the electron . v = 4 x 105 m/s _____________________________________________________________ solution: (a) magnetic field (magnitude and direction) at points, P1, P2 and P3 is , at : P1 is , B1 = o I1 / 2r1 = (4x10-7 A/m )(4 A) / 2(0.01 m) = 800 x10-7 T (in to page ) at : P2 is , B2 = o I1 / 2r1 - o I2 / 2r1 = 800 x10-7 T - 600 x10-7 T = 200 x10-7 T (out of page) at : P3 is ,B3 =o I2 / 2r3 = 800 x10-7 T (in to page) __________________________________________________________ ___________________________________________________________ (b) acceleration (magnitude and direction) of an electronis , caculated by applying newton's law as , ma = q v B2 a = q v B2 /m = (1.6 x10-19 C)(4 x 105 m/s)(200 x10-7 T )/ (9.1 x10-31kg ) = 140.6 x1010 m/s (out of page) _______________________________________________________________ ________________________________________________________________ (c) In order to find the force (magnitude and direction) that the left wire exerts on the right is given by , F = o I1 I2 / 2(r1 + d + r3) = (4x10-7 A/m )(3A )(4A ) / (0.04 m) = 600 N (right) ______________________________________________________________ ______________________________________________________________ (d) x-axis magnetic field will be vanished on x axis at when , o I1 / 2r1 + o I2 / 2 (d-x) = o I2 / 2r3 Here , x is the point where field is 0 I1 / r1 + I2 / (d-x) = I2/ r3 4 A / (0.01 m) + 3A / (0.02 -x )= (3 A / 0.01m ) 400 A/m + (3 A / 0.02 -x ) = 300 A/m 100 A/m = 3 A / 0.02 -x 0.02 -x = 0.03 m x = - 0.01 m Hence , from I2 of left at a distance of 0.01 m field should be vanshed
currents in the wires , 1 and 2 is I1 = 4 A I2 = 3A distance between the wires , d = 2 cm = 2 cm (10-2 m /1 cm ) = 0.02 m distance b/w current I1 to point P1 is , r 1 = 1 cm = 1 cm (10-2 m /1 cm ) = 0.01 m distance b/w current I2 to point P3 is , r 3 = 1 cm = 1 cm (10-2 m /1 cm ) = 0.01 m distance b/w I1 or I2 to piont P2 is , r 2 = 2/2 cm = 1 cm (10-2 m /1 cm ) = 0.01 m velocity of the electron . v = 4 x 105 m/s _____________________________________________________________ solution: (a) magnetic field (magnitude and direction) at points, P1, P2 and P3 is , at : P1 is , B1 = o I1 / 2r1 = (4x10-7 A/m )(4 A) / 2(0.01 m) = 800 x10-7 T (in to page ) at : P2 is , B2 = o I1 / 2r1 - o I2 / 2r1 = 800 x10-7 T - 600 x10-7 T = 200 x10-7 T (out of page) at : P3 is ,B3 =o I2 / 2r3 = 800 x10-7 T (in to page) __________________________________________________________ ___________________________________________________________ (b) acceleration (magnitude and direction) of an electron
is , caculated by applying newton's law as , ma = q v B2 a = q v B2 /m = (1.6 x10-19 C)(4 x 105 m/s)(200 x10-7 T )/ (9.1 x10-31kg ) = 140.6 x1010 m/s (out of page) _______________________________________________________________ ________________________________________________________________ (c) In order to find the force (magnitude and direction) that the left wire exerts on the right is given by , F = o I1 I2 / 2(r1 + d + r3) = (4x10-7 A/m )(3A )(4A ) / (0.04 m) = 600 N (right) ______________________________________________________________ ______________________________________________________________ (d) x-axis magnetic field will be vanished on x axis at when , o I1 / 2r1 + o I2 / 2 (d-x) = o I2 / 2r3 Here , x is the point where field is 0 I1 / r1 + I2 / (d-x) = I2/ r3 4 A / (0.01 m) + 3A / (0.02 -x )= (3 A / 0.01m ) 400 A/m + (3 A / 0.02 -x ) = 300 A/m 100 A/m = 3 A / 0.02 -x 0.02 -x = 0.03 m x = - 0.01 m Hence , from I2 of left at a distance of 0.01 m field should be vanshed
= 0.01 m distance b/w current I2 to point P3 is , r 3 = 1 cm = 1 cm (10-2 m /1 cm ) = 0.01 m distance b/w I1 or I2 to piont P2 is , r 2 = 2/2 cm = 1 cm (10-2 m /1 cm ) = 0.01 m velocity of the electron . v = 4 x 105 m/s _____________________________________________________________ solution: (a) magnetic field (magnitude and direction) at points, P1, P2 and P3 is , at : P1 is , B1 = o I1 / 2r1 = (4x10-7 A/m )(4 A) / 2(0.01 m) = 800 x10-7 T (in to page ) at : P2 is , B2 = o I1 / 2r1 - o I2 / 2r1 = 800 x10-7 T - 600 x10-7 T = 200 x10-7 T (out of page) at : P3 is ,B3 =o I2 / 2r3 = 800 x10-7 T (in to page) __________________________________________________________ ___________________________________________________________ (b) acceleration (magnitude and direction) of an electron
is , caculated by applying newton's law as , ma = q v B2 a = q v B2 /m = (1.6 x10-19 C)(4 x 105 m/s)(200 x10-7 T )/ (9.1 x10-31kg ) = 140.6 x1010 m/s (out of page) _______________________________________________________________ ________________________________________________________________ (c) In order to find the force (magnitude and direction) that the left wire exerts on the right is given by , F = o I1 I2 / 2(r1 + d + r3) = (4x10-7 A/m )(3A )(4A ) / (0.04 m) = 600 N (right) ______________________________________________________________ ______________________________________________________________ (d) x-axis magnetic field will be vanished on x axis at when , o I1 / 2r1 + o I2 / 2 (d-x) = o I2 / 2r3 Here , x is the point where field is 0 I1 / r1 + I2 / (d-x) = I2/ r3 4 A / (0.01 m) + 3A / (0.02 -x )= (3 A / 0.01m ) 400 A/m + (3 A / 0.02 -x ) = 300 A/m 100 A/m = 3 A / 0.02 -x 0.02 -x = 0.03 m x = - 0.01 m Hence , from I2 of left at a distance of 0.01 m field should be vanshed
r 3 = 1 cm = 1 cm (10-2 m /1 cm ) = 0.01 m distance b/w I1 or I2 to piont P2 is , r 2 = 2/2 cm = 1 cm (10-2 m /1 cm ) = 0.01 m velocity of the electron . v = 4 x 105 m/s _____________________________________________________________ solution: (a) magnetic field (magnitude and direction) at points, P1, P2 and P3 is , at : P1 is , B1 = o I1 / 2r1 = (4x10-7 A/m )(4 A) / 2(0.01 m) = 800 x10-7 T (in to page ) at : P2 is , B2 = o I1 / 2r1 - o I2 / 2r1 = 800 x10-7 T - 600 x10-7 T = 200 x10-7 T (out of page) at : P3 is ,B3 =o I2 / 2r3 = 800 x10-7 T (in to page) __________________________________________________________ ___________________________________________________________ (b) acceleration (magnitude and direction) of an electron
is , caculated by applying newton's law as , ma = q v B2 a = q v B2 /m = (1.6 x10-19 C)(4 x 105 m/s)(200 x10-7 T )/ (9.1 x10-31kg ) = 140.6 x1010 m/s (out of page) _______________________________________________________________ ________________________________________________________________ (c) In order to find the force (magnitude and direction) that the left wire exerts on the right is given by , F = o I1 I2 / 2(r1 + d + r3) = (4x10-7 A/m )(3A )(4A ) / (0.04 m) = 600 N (right) ______________________________________________________________ ______________________________________________________________ (d) x-axis magnetic field will be vanished on x axis at when , o I1 / 2r1 + o I2 / 2 (d-x) = o I2 / 2r3 Here , x is the point where field is 0 I1 / r1 + I2 / (d-x) = I2/ r3 4 A / (0.01 m) + 3A / (0.02 -x )= (3 A / 0.01m ) 400 A/m + (3 A / 0.02 -x ) = 300 A/m 100 A/m = 3 A / 0.02 -x 0.02 -x = 0.03 m x = - 0.01 m Hence , from I2 of left at a distance of 0.01 m field should be vanshed
= 0.01 m distance b/w I1 or I2 to piont P2 is , r 2 = 2/2 cm = 1 cm (10-2 m /1 cm ) = 0.01 m velocity of the electron . v = 4 x 105 m/s _____________________________________________________________ solution: (a) magnetic field (magnitude and direction) at points, P1, P2 and P3 is , at : P1 is , B1 = o I1 / 2r1 = (4x10-7 A/m )(4 A) / 2(0.01 m) = 800 x10-7 T (in to page ) at : P2 is , B2 = o I1 / 2r1 - o I2 / 2r1 = 800 x10-7 T - 600 x10-7 T = 200 x10-7 T (out of page) at : P3 is ,B3 =o I2 / 2r3 = 800 x10-7 T (in to page) __________________________________________________________ ___________________________________________________________ (b) acceleration (magnitude and direction) of an electron
is , caculated by applying newton's law as , ma = q v B2 a = q v B2 /m = (1.6 x10-19 C)(4 x 105 m/s)(200 x10-7 T )/ (9.1 x10-31kg ) = 140.6 x1010 m/s (out of page) _______________________________________________________________ ________________________________________________________________ (c) In order to find the force (magnitude and direction) that the left wire exerts on the right is given by , F = o I1 I2 / 2(r1 + d + r3) = (4x10-7 A/m )(3A )(4A ) / (0.04 m) = 600 N (right) ______________________________________________________________ ______________________________________________________________ (d) x-axis magnetic field will be vanished on x axis at when , o I1 / 2r1 + o I2 / 2 (d-x) = o I2 / 2r3 Here , x is the point where field is 0 I1 / r1 + I2 / (d-x) = I2/ r3 4 A / (0.01 m) + 3A / (0.02 -x )= (3 A / 0.01m ) 400 A/m + (3 A / 0.02 -x ) = 300 A/m 100 A/m = 3 A / 0.02 -x 0.02 -x = 0.03 m x = - 0.01 m Hence , from I2 of left at a distance of 0.01 m field should be vanshed
Here , x is the point where field is 0 I1 / r1 + I2 / (d-x) = I2/ r3 4 A / (0.01 m) + 3A / (0.02 -x )= (3 A / 0.01m ) 400 A/m + (3 A / 0.02 -x ) = 300 A/m 100 A/m = 3 A / 0.02 -x 0.02 -x = 0.03 m x = - 0.01 m Hence , from I2 of left at a distance of 0.01 m field should be vanshed 4 A / (0.01 m) + 3A / (0.02 -x )= (3 A / 0.01m ) 400 A/m + (3 A / 0.02 -x ) = 300 A/m 100 A/m = 3 A / 0.02 -x 0.02 -x = 0.03 m x = - 0.01 m Hence , from I2 of left at a distance of 0.01 m field should be vanshed
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.