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Question: Shown in the figure is a circular loop of radius $10^{-3}$ m has its plane parallel to the wire and ...

Shown in the figure is a circular loop of radius 10−310^{-3} m has its plane parallel to the wire and its centre is placed at a distance of 1m from the wire. The resistance of the loop is 8.4×10−4Ω8.4 \times 10^{-4} \Omega. Find the magnitude and the direction of the induced current in the loop.

Answer

Magnitude: 7.48×10−77.48 \times 10^{-7} A Direction: Clockwise

Explanation

Solution

The problem describes a circular loop moving away from a long straight current-carrying wire. We need to find the induced current in the loop.

1. Understand the Geometry and Magnetic Field: Let the long straight wire carry current IwI_w along the y-axis. The magnetic field produced by this wire at a perpendicular distance rr is given by B=μ0Iw2πrB = \frac{\mu_0 I_w}{2\pi r}. For the loop to experience a changing magnetic flux, its plane must be perpendicular to the magnetic field lines. If the wire is along the y-axis, the magnetic field lines are circles in the x-z plane. Thus, the loop must be in the x-y plane (or any plane parallel to it, like the x-z plane if the wire is along y-axis). The phrase "plane parallel to the wire" implies that the loop's plane contains lines parallel to the wire. For a wire along the y-axis, this means the loop is in the x-y plane. In this configuration, the magnetic field is perpendicular to the plane of the loop (along the z-axis).

Let the center of the circular loop be at a distance xx from the wire. The magnetic field at a point within the loop at a distance rr from the wire is B(r)=μ0Iw2πrB(r) = \frac{\mu_0 I_w}{2\pi r}.

2. Magnetic Flux through the Loop: The radius of the loop is a=10−3a = 10^{-3} m, and its center is at a distance x=1x = 1 m from the wire. Since a≪xa \ll x (10−3≪110^{-3} \ll 1), we can approximate the magnetic field as uniform over the area of the loop and equal to the field at its center. The magnetic field at the center of the loop is B(x)=μ0Iw2πxB(x) = \frac{\mu_0 I_w}{2\pi x}. The area of the loop is A=πa2A = \pi a^2. The magnetic flux ΦB\Phi_B through the loop is approximately: ΦB=B(x)⋅A=(μ0Iw2πx)(πa2)=μ0Iwa22x\Phi_B = B(x) \cdot A = \left(\frac{\mu_0 I_w}{2\pi x}\right) (\pi a^2) = \frac{\mu_0 I_w a^2}{2x}

3. Induced Electromotive Force (EMF): The loop is moving away from the wire with velocity v=10v = 10 m/s. This means the distance xx is changing with time, i.e., dxdt=v\frac{dx}{dt} = v. According to Faraday's Law of Induction, the induced EMF is E=−dΦBdt\mathcal{E} = -\frac{d\Phi_B}{dt}. E=−ddt(μ0Iwa22x)\mathcal{E} = -\frac{d}{dt} \left( \frac{\mu_0 I_w a^2}{2x} \right) Since μ0,Iw,a\mu_0, I_w, a are constants: E=−μ0Iwa22ddt(1x)=−μ0Iwa22(−1x2dxdt)\mathcal{E} = -\frac{\mu_0 I_w a^2}{2} \frac{d}{dt} \left( \frac{1}{x} \right) = -\frac{\mu_0 I_w a^2}{2} \left( -\frac{1}{x^2} \frac{dx}{dt} \right) E=μ0Iwa2v2x2\mathcal{E} = \frac{\mu_0 I_w a^2 v}{2x^2} Now, substitute the given values: μ0=4π×10−7\mu_0 = 4\pi \times 10^{-7} T m/A Iw=100I_w = 100 A a=10−3a = 10^{-3} m v=10v = 10 m/s x=1x = 1 m E=(4π×10−7 T m/A)(100 A)(10−3 m)2(10 m/s)2(1 m)2\mathcal{E} = \frac{(4\pi \times 10^{-7} \text{ T m/A}) (100 \text{ A}) (10^{-3} \text{ m})^2 (10 \text{ m/s})}{2 (1 \text{ m})^2} E=4π×10−7×102×10−6×102\mathcal{E} = \frac{4\pi \times 10^{-7} \times 10^2 \times 10^{-6} \times 10}{2} E=4π×10−7+2−6+12=4π×10−102=2π×10−10 V\mathcal{E} = \frac{4\pi \times 10^{-7+2-6+1}}{2} = \frac{4\pi \times 10^{-10}}{2} = 2\pi \times 10^{-10} \text{ V}

4. Induced Current: The resistance of the loop is R=8.4×10−4ΩR = 8.4 \times 10^{-4} \Omega. Using Ohm's Law, the induced current IinducedI_{induced} is: Iinduced=ER=2π×10−10 V8.4×10−4ΩI_{induced} = \frac{\mathcal{E}}{R} = \frac{2\pi \times 10^{-10} \text{ V}}{8.4 \times 10^{-4} \Omega} Iinduced=2π8.4×10−6 A=π4.2×10−6 AI_{induced} = \frac{2\pi}{8.4} \times 10^{-6} \text{ A} = \frac{\pi}{4.2} \times 10^{-6} \text{ A} Using π≈3.14159\pi \approx 3.14159: Iinduced=3.141594.2×10−6 A≈0.74799×10−6 AI_{induced} = \frac{3.14159}{4.2} \times 10^{-6} \text{ A} \approx 0.74799 \times 10^{-6} \text{ A} Iinduced≈7.48×10−7 AI_{induced} \approx 7.48 \times 10^{-7} \text{ A}

5. Direction of Induced Current (Lenz's Law): Assume the current in the wire is upwards. By the right-hand rule, the magnetic field produced by the wire to its right (where the loop is located) points into the plane of the loop. As the loop moves away from the wire, the distance xx increases. Since B∝1/xB \propto 1/x, the magnetic field strength through the loop decreases. This means the magnetic flux into the page through the loop is decreasing. According to Lenz's Law, the induced current will flow in a direction that opposes this change. To oppose the decrease in flux into the page, the induced current must create its own magnetic field pointing into the page. By the right-hand rule for a loop, a current flowing in a clockwise direction produces a magnetic field into the page. Therefore, the induced current is clockwise.

The final answer is 7.48×10−7 A, clockwise\boxed{7.48 \times 10^{-7} \text{ A, clockwise}}.