Question

A square loop of side 'a' & resistance R moves with a uniform velocity v away from a long wire that carries current I as shown in figure. The loop is moved away from the wire with side AB always parallel to the wire. Initially, distance between the side AB of the loop & wire is 'a' The work done when the loop is moved through a distance 'a' from initial position is $\Delta w = \frac{\mu_0^2I^2av}{2\pi^2R} [\frac{1}{3} + ln(k)]$. Find the value of k.

Answer 1

0.75

Answer 2

The problem asks us to find the value of 'k' by comparing the calculated work done with the given expression for work done.

1. Magnetic Field and Flux: The magnetic field produced by a long straight wire carrying current I at a distance 'x' is given by: $B(x) = \frac{\mu_0 I}{2\pi x}$

The magnetic flux ($\Phi$) through the square loop of side 'a' when its inner side (AB) is at a distance 'x' from the wire can be calculated by integrating the magnetic field over the area of the loop.

Consider an elemental strip of width 'dr' at a distance 'r' from the wire. The area of this strip is $dA = a \cdot dr$. The flux through this elemental strip is $d\Phi = B(r) dA = \frac{\mu_0 I}{2\pi r} a dr$.

The total flux through the loop is obtained by integrating from $x$ (distance of side AB) to $x+a$ (distance of side CD): $\Phi = \int_x^{x+a} \frac{\mu_0 I a}{2\pi r} dr = \frac{\mu_0 I a}{2\pi} [\ln r]_x^{x+a}$ $\Phi = \frac{\mu_0 I a}{2\pi} \left[\ln(x+a) - \ln(x)\right] = \frac{\mu_0 I a}{2\pi} \ln\left(\frac{x+a}{x}\right)$

2. Induced EMF: As the loop moves with a uniform velocity 'v' away from the wire, 'x' changes with time ($dx/dt = v$). The induced EMF ($\mathcal{E}$) is given by Faraday's law of electromagnetic induction: $\mathcal{E} = -\frac{d\Phi}{dt} = -\frac{d\Phi}{dx} \frac{dx}{dt} = -v \frac{d\Phi}{dx}$ $\mathcal{E} = -v \frac{d}{dx}\left(\frac{\mu_0 I a}{2\pi} \ln\left(\frac{x+a}{x}\right)\right)$ $\mathcal{E} = -v \frac{\mu_0 I a}{2\pi} \frac{d}{dx}\left(\ln(x+a) - \ln(x)\right)$ $\mathcal{E} = -v \frac{\mu_0 I a}{2\pi} \left(\frac{1}{x+a} - \frac{1}{x}\right)$ $\mathcal{E} = -v \frac{\mu_0 I a}{2\pi} \left(\frac{x - (x+a)}{x(x+a)}\right) = -v \frac{\mu_0 I a}{2\pi} \left(\frac{-a}{x(x+a)}\right)$ $\mathcal{E} = \frac{\mu_0 I a^2 v}{2\pi x(x+a)}$

3. Induced Current: The induced current ($i$) in the loop, given its resistance R, is: $i = \frac{\mathcal{E}}{R} = \frac{\mu_0 I a^2 v}{2\pi R x(x+a)}$

4. Force on the Loop: The magnetic force on the loop opposes its motion (Lenz's Law). The forces on sides AD and BC cancel out. The forces on sides AB and CD are:

Force on AB ($F_{AB}$): This side is closer to the wire, so the field is stronger. The current in AB (clockwise direction, to induce a field into the page to oppose the decrease in flux) will be upwards. Force $F_{AB} = i a B_{AB}$ (to the left, attractive). $F_{AB} = \left(\frac{\mu_0 I a^2 v}{2\pi R x(x+a)}\right) a \left(\frac{\mu_0 I}{2\pi x}\right) = \frac{\mu_0^2 I^2 a^3 v}{4\pi^2 R x^2(x+a)}$

Force on CD ($F_{CD}$): The current in CD will be downwards. Force $F_{CD} = i a B_{CD}$ (to the right, repulsive). $F_{CD} = \left(\frac{\mu_0 I a^2 v}{2\pi R x(x+a)}\right) a \left(\frac{\mu_0 I}{2\pi (x+a)}\right) = \frac{\mu_0^2 I^2 a^3 v}{4\pi^2 R x(x+a)^2}$

The net magnetic force opposing the motion (to the left) is $F_{net} = F_{AB} - F_{CD}$: $F_{net} = \frac{\mu_0^2 I^2 a^3 v}{4\pi^2 R} \left( \frac{1}{x^2(x+a)} - \frac{1}{x(x+a)^2} \right)$ $F_{net} = \frac{\mu_0^2 I^2 a^3 v}{4\pi^2 R} \left( \frac{(x+a) - x}{x^2(x+a)^2} \right) = \frac{\mu_0^2 I^2 a^3 v}{4\pi^2 R} \frac{a}{x^2(x+a)^2}$ $F_{net} = \frac{\mu_0^2 I^2 a^4 v}{4\pi^2 R x^2(x+a)^2}$

To move the loop with uniform velocity 'v', an external force equal in magnitude and opposite in direction to $F_{net}$ must be applied. So, $F_{ext} = F_{net}$.

5. Work Done: The work done ($\Delta w$) when the loop is moved through a distance 'a' from its initial position (where the distance between AB and the wire is 'a') is: $\Delta w = \int_{initial\_x}^{final\_x} F_{ext} dx$

Initial position: $x_1 = a$. Final position: $x_2 = a+a = 2a$. $\Delta w = \int_a^{2a} \frac{\mu_0^2 I^2 a^4 v}{4\pi^2 R x^2(x+a)^2} dx$

The constant term is $\frac{\mu_0^2 I^2 a^4 v}{4\pi^2 R}$. We need to evaluate the integral: $J = \int_a^{2a} \frac{1}{x^2(x+a)^2} dx$

Using partial fraction decomposition: $\frac{1}{x^2(x+a)^2} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+a} + \frac{D}{(x+a)^2}$ Multiplying by $x^2(x+a)^2$: $1 = Ax(x+a)^2 + B(x+a)^2 + Cx^2(x+a) + Dx^2$

Setting $x=0 \implies 1 = B(a^2) \implies B = 1/a^2$. Setting $x=-a \implies 1 = D(-a)^2 \implies D = 1/a^2$. Comparing coefficients of $x^3$: $A+C=0 \implies C=-A$. Comparing coefficients of $x^2$: $A(2a) + B + C(a) + D = 0 \implies 2aA + 1/a^2 + aC + 1/a^2 = 0$ $2aA + aC + 2/a^2 = 0$. Substitute $C=-A$: $2aA - aA + 2/a^2 = 0 \implies aA + 2/a^2 = 0 \implies A = -2/a^3$. So, $C = 2/a^3$.

Thus, the integrand is: $\frac{1}{x^2(x+a)^2} = -\frac{2}{a^3 x} + \frac{1}{a^2 x^2} + \frac{2}{a^3 (x+a)} + \frac{1}{a^2 (x+a)^2}$

Now, integrate: $J = \left[ -\frac{2}{a^3} \ln|x| - \frac{1}{a^2 x} + \frac{2}{a^3} \ln|x+a| - \frac{1}{a^2 (x+a)} \right]_a^{2a}$ $J = \left[ \frac{2}{a^3} \ln\left|\frac{x+a}{x}\right| - \frac{1}{a^2 x} - \frac{1}{a^2 (x+a)} \right]_a^{2a}$

Evaluate at the upper limit ($x=2a$): $\frac{2}{a^3} \ln\left(\frac{2a+a}{2a}\right) - \frac{1}{a^2 (2a)} - \frac{1}{a^2 (2a+a)} = \frac{2}{a^3} \ln\left(\frac{3}{2}\right) - \frac{1}{2a^3} - \frac{1}{3a^3}$ $= \frac{2}{a^3} \ln\left(\frac{3}{2}\right) - \frac{3+2}{6a^3} = \frac{2}{a^3} \ln\left(\frac{3}{2}\right) - \frac{5}{6a^3}$

Evaluate at the lower limit ($x=a$): $\frac{2}{a^3} \ln\left(\frac{a+a}{a}\right) - \frac{1}{a^2 (a)} - \frac{1}{a^2 (a+a)} = \frac{2}{a^3} \ln(2) - \frac{1}{a^3} - \frac{1}{2a^3}$ $= \frac{2}{a^3} \ln(2) - \frac{2+1}{2a^3} = \frac{2}{a^3} \ln(2) - \frac{3}{2a^3}$

Subtract the lower limit from the upper limit: $J = \left( \frac{2}{a^3} \ln\left(\frac{3}{2}\right) - \frac{5}{6a^3} \right) - \left( \frac{2}{a^3} \ln(2) - \frac{3}{2a^3} \right)$ $J = \frac{2}{a^3} \left( \ln\left(\frac{3}{2}\right) - \ln(2) \right) - \frac{5}{6a^3} + \frac{3}{2a^3}$ $J = \frac{2}{a^3} \ln\left(\frac{3/2}{2}\right) + \frac{-5+9}{6a^3}$ $J = \frac{2}{a^3} \ln\left(\frac{3}{4}\right) + \frac{4}{6a^3} = \frac{2}{a^3} \ln\left(\frac{3}{4}\right) + \frac{2}{3a^3}$

Now, substitute J back into the work done expression: $\Delta w = \frac{\mu_0^2 I^2 a^4 v}{4\pi^2 R} \left( \frac{2}{a^3} \ln\left(\frac{3}{4}\right) + \frac{2}{3a^3} \right)$ $\Delta w = \frac{\mu_0^2 I^2 a v}{4\pi^2 R} \left( 2 \ln\left(\frac{3}{4}\right) + \frac{2}{3} \right)$

Factor out 2 from the bracket and simplify the leading constant: $\Delta w = \frac{\mu_0^2 I^2 a v}{2\pi^2 R} \left( \ln\left(\frac{3}{4}\right) + \frac{1}{3} \right)$

6. Compare with the given expression: The given work done is $\Delta w = \frac{\mu_0^2I^2av}{2\pi^2R} [\frac{1}{3} + ln(k)]$. Comparing our calculated expression with the given one: $\frac{\mu_0^2I^2av}{2\pi^2R} [\frac{1}{3} + \ln(\frac{3}{4})] = \frac{\mu_0^2I^2av}{2\pi^2R} [\frac{1}{3} + ln(k)]$ Therefore, $\ln(k) = \ln(\frac{3}{4})$, which implies $k = \frac{3}{4}$.

The final answer is $\boxed{0.75}$.