Answer:
The magnitude of the magnetic field made by current in the wire is 3.064 x 10⁻⁶ T.
Explanation:
Given;
length of the straight wire, L = 0.56 m
conventional current, I = 0.4 A
distance of magnetic field from the wire, r = 2.6 cm = 0.026 m
To determine magnitude of magnetic field made by current in the wire, we will apply Bio-Savart Law;

Therefore, the magnitude of the magnetic field made by current in the wire is 3.064 x 10⁻⁶ T.
Answer:
0.21486 mm
Explanation:
The formula for the maximum intensity is given by;
I = I_o•cos²(Φ/2)
Now,we are not given Φ but it can be expressed in terms of what we are given as; Φ = πdy/(λL)
Where;
y is the distance from the central maximum
d is the distance between the slits
λ is the wavelength
L is the distance to the screen
Thus;
I = I_o•πdy/(λL)
We are given;
d = 0.05 mm = 0.5 × 10^(-3) m
λ = 540 nm = 540 × 10^(-9) m
L = 1.25 m
I/I_o = 50% = 0.5
From earlier, we saw that;
I = I_o•πdy/(λL)
We have I/I_o = 0.5
Thus;
I/I_o = πdy/(λL)
Plugging in the relevant values;
0.5 = (π × 0.5 × 10^(-3) × y)/(540 × 10^(-9) × 1.25)
Making y the subject, we have;
y = (0.5 × 540 × 10^(-9) × 1.25)/(π × 0.5 × 10^(-3))
y = 0.00021486 m
Converting to mm, we have;
y = 0.21486 mm
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Answer:
<em>The direction of the magnetic field on point P, equidistant from both wires, and having equal magnitude of current flowing through them will be pointed perpendicularly away from the direction of the wires.</em>
Explanation:
Using the right hand grip, the direction of the magnet field on the wire M is counterclockwise, and the direction of the magnetic field on wire N is clockwise. Using this ideas, we can see that the magnetic flux of both field due to the currents of the same magnitude through both wires, acting on a particle P equidistant from both wires will act in a direction perpendicularly away from both wires.