A long, straight wire of length 2L on the y-axis carries a current I. According to the Biot-Savart Law, the magnitude of the magnetic field due to the current at a point (a,0) is given by
B(a)= (mu sub 0 multiply by I) divided by 4pi times the integral from -L to L of (Sine beta / r^2)dy
where mu sub 0 is a physical constant, a>0, and beta, r, and y are related.
a)Show that the magnitude of the magnetic field at (a,0) is
B(a)= (mu sub 0 times I and L) divided by (2pi*a*sqrt(a^2 + L^2))
b)what is the magnitude of the magnetic field at (a,0) due to an infinitely long wire
(L -> infinity)?
Note: (*) means multiply by
To solve this problem, we will use the Biot-Savart Law, which provides a method for calculating the magnetic field generated by a current-carrying wire at a specific point.
a) To compute the magnitude of the magnetic field at point (a,0), we start by expressing the given integral in terms of the variables given in the problem: beta, r, and y.
The integral in the Biot-Savart Law is:
B(a) = (mu sub 0 * I) / (4pi) * ∫[-L to L] (sin(beta) / r^(2)) dy
Here, beta and r are related to the coordinates of the point (a,0). From the diagram given, we can see that beta is the angle between r (the distance from the wire element to point (a,0)) and the y-axis. Additionally, r and y are related by r = sqrt(a^2 + y^2).
Taking these relationships into account, we can rewrite the integral as follows:
B(a) = (mu sub 0 * I) / (4pi) * ∫[-L to L] (sin(beta) / (a^2 + y^2)) dy
Next, we need to express sin(beta) in terms of a, y, and r. From the right triangle formed by the wire element, point (a,0), and point (a,y), we know that sin(beta) = y / r.
Substituting this expression into the integral, we have:
B(a) = (mu sub 0 * I) / (4pi) * ∫[-L to L] (y / (r * (a^2 + y^2))) dy
Now, we can substitute r = sqrt(a^2 + y^2) into the integral:
B(a) = (mu sub 0 * I) / (4pi) * ∫[-L to L] (y / (sqrt(a^2 + y^2) * (a^2 + y^2))) dy
To evaluate this integral, we can use a substitution. Let u = a^2 + y^2, so du = 2y dy. Substituting these values, we get:
B(a) = (mu sub 0 * I) / (4pi) * 1/2 * ∫[-u=L^2 + a^2 to u=L^2 + a^2] (1 / (sqrt(u) * u)) du
Simplifying the integral, we obtain:
B(a) = (mu sub 0 * I) / (8pi) * ∫[-u=L^2 + a^2 to u=L^2 + a^2] (u^(-3/2)) du
Integrating the expression u^(-3/2), we find:
B(a) = (mu sub 0 * I) / (8pi) * [-2u^(-1/2)] evaluated at u=L^2 + a^2 and u=L^2 + a^2
Plugging in these values, we have:
B(a) = (mu sub 0 * I) / (8pi) * [-2/(L^2 + a^2)^(-1/2) - 2/(L^2 + a^2)^(-1/2)]
Simplifying further, we get:
B(a) = (mu sub 0 * I) / (8pi) * [-4/(L^2 + a^2)^(-1/2)]
Using the formula for the square root of a sum of squares, we rewrite the expression as:
B(a) = (mu sub 0 * I) / (8pi) * (-4 / sqrt((a^2 + L^2)^2))
Finally, simplifying the expression yields:
B(a) = (mu sub 0 * I) / (2pi) * (1 / sqrt(a^2 + L^2))
b) To determine the magnitude of the magnetic field at point (a,0) due to an infinitely long wire (L -> infinity), we can take the limit of the expression from part a as L approaches infinity.
lim(L -> infinity) B(a) = (mu sub 0 * I) / (2pi) * (1 / sqrt(a^2 + L^2))
As L becomes very large, we can ignore the a^2 in the denominator compared to L^2. Therefore, the expression simplifies to:
lim(L -> infinity) B(a) = (mu sub 0 * I) / (2pi * L)
Hence, in the case of an infinitely long wire, the magnitude of the magnetic field at point (a,0) is (mu sub 0 * I) divided by (2pi * L).