A wire carries a 7.00-A current along the x-axis, and another wire carries a6.00

Question

A wire carries a 7.00-A current along the x-axis, and another wire carries a6.00-A current along the y-axis, as shown in Fig-2. What is the net magnetic field at point P, located at x = 5 4.00 m, y = 53.00 m? In Fig-3 the battery emf is 50.0 V, the resistance is 250 ohm, and capacitance is 0.50 mu F. The switch S is closed for a long time interval, and zero difference is measured across the capacitor. After the switch is opened, the potential difference across the capacitor reaches a maximum values of 150 V. What is the inductance?

Explanation / Answer

If both currents are in the positive axis directions, then their magnetic fields will be in opposite directions and will therefore oppose each other. Assuming conventional current, the field will be OUT of the page from the x-axis current, and IN to the page from the y-axis current. Use the right-hand rule

Magnetic field at P(54,53) due to x-axis current (53m away from x-axis) is

B = u0*7A/2pi*53

B = (4pi x 10^-7 x 7) / (2pi x 53)

B = 0. 26415 x 10^-7 T ...............OUT of page

Magnetic field at P(54,53) due to x-axis current (54m away from x-axis) is

B = u0*7A/2pi*54

B = (4pi x 10^-7 x 7) / (2pi x 54)

B = 0.25926 x 10^-7 T ...............IN to the page

So net B field at P(54,53) is:

(0.26415 - 0.25926) x 10^-7 T = 0.00489 x 10^-7 T = 4.89 x 10^-10 T = 4.89 nT

out of page, perpendicular to y and x axes

for problem 2. figure is not given..

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