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jeka57 [31]
3 years ago
14

A long wire carrying a 5.0 A current perpendicular to the xy-plane intersects the x-axis at x= - 2.0 cm . A second, parallel wir

e carrying a 3.5 A current intersects the x-axis at x=+ 2.0cm .
A). At what point on the x-axis is the magnetic field zero if the two currents are in the same direction?

B). At what point on the x-axis is the magnetic field zero if the two currents are in opposite directions?
Physics
1 answer:
mario62 [17]3 years ago
8 0

Answer:

a . 0.35cm

b.  11.33cm

Explanation:

a. Given both currents are in the same direction, the null point lies in between them. Let x be distance of N from first wire, then distance from 2nd wire is 4-x

#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in between the wires:

\frac{\mu_oi_1}{2\pi x}=\frac{\mu_oi_2}{2\pi(4-x)}\\\\5/x=\frac{3.5}{4-x}\\\\x=2.35cm\\\\N=2.35-2=0.35cm

Hence, for currents in same direction, the point is 0.35cm

b. Given both currents flow in opposite directions, the null point lies on the other side.

#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in outside the wires:

Let x be distance of N from first wire, then distance from 2nd wire is 4+x:

\frac{\mu_oi_1}{2\pi(4+ x)}=\frac{\mu_oi_2}{2\pi x}\\\\5/(4+x)=\frac{3.5}{x}\\\\x=9.33cm\\\\N=9.33+2=11.33cm

Hence, if currents are in opposite directions the point on x-axis is 11.33cm

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3 years ago
A block of mass 10 kg and measuring 250 mm on each edge is pulled up an inclined surface on which there is a film of SAE 10W-30
Ivan

Answer:

a) 2.53 * 10^-2 m/s

b) -4.78 * 10^-2 m/s

c)  1.21 * 10^-1 m/s

Explanation:

Given data :

Mass of block = 10 kg

Measuring 250mm on each side

a) calculate the speed  when a force of 75N is applied to pull block upwards

F = f + W sin∅ ( equation for applying the force of equilibrium condition in the x axis )  ----- ( 1 )

f ( friction force )=  ( 16400v * 6.25 *10^-2) =  1025 v

F ( force applied ) = 75

W ( weight of  block ) = 10 * 9.81 = 98.1 N

∅ = 30°

input values into equation 1

V = \frac{75- (98.1*sin30^{0}) }{1025} = 2.53 * 10^-2 m/s

b) Speed when no force is applied on the block

F = f + W sin∅

F = 0

f = 1025 V

W = 98.1 N

∅ = 30°

hence V = \frac{0 - (98.1*sin30^{0}) }{1025} =  - 4.78 * 10^-2 m/s

c) when a force is applied to push block down the incline

F = f + W sin∅ ----- ( 3 )

F = 75 N

f = 1025 V

W = 98.1 N

∅ = 30°

input values into equation 3 considering the fact that the weight of the block is acting in the opposite direction

75 = 1025 V - 98.1 ( sin 30° )

V = \frac{75+( 98.1*sin30^{0})  }{1025} = 1.21 * 10^-1 m/s

5 0
3 years ago
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