The long straight wire in the figure has current I = 1 A flowing in it.A square loop which has 10-cm sides is positioned 10 cm away from the wire as shown.The loop is then moved in the positive x-direction with a speed v = 10 cm/s.If the loop has a resistance of 0.02 ohms,calculate the direction and the magnitude of the induced current in the loop the instant the loop is made to move.
A)0.50 $\mu$A clockwise
B)0.50 $\mu$A counter-clockwise
C)6.7 $\mu$A clockwise
D)6.7 $\mu$A counter-clockwise
E)1.3 $\mu$A clockwise

Question

The long straight wire in the figure has current I = 1 A flowing in it.A square loop which has 10-cm sides is positioned 10 cm away from the wire as shown.The loop is then moved in the positive x-direction with a speed v = 10 cm/s.If the loop has a resistance of 0.02 ohms,calculate the direction and the magnitude of the induced current in the loop the instant the loop is made to move. 
A)0.50 $\mu$A clockwise 
B)0.50 $\mu$A counter-clockwise 
C)6.7 $\mu$A clockwise 
D)6.7 $\mu$A counter-clockwise 
E)1.3 $\mu$A clockwise

Quizplus · Accepted Answer

When the loop moves in the positive x-direction, the magnetic field due to the wire passing through the loop will also change in the positive x-direction. Using the right-hand rule, we can determine that the induced current in the loop will create a magnetic field in the opposite direction to the change in the external magnetic field. Therefore, the induced current will flow in a clockwise direction. Applying Faraday's Law, we can calculate the magnitude of the induced emf as:
$$
\epsilon = -\frac{d\Phi}{dt}
$$
where $\Phi$ is the magnetic flux through the loop. At the instant the loop starts moving, the magnetic flux through the loop is given by $\Phi = BA$, where $B$ is the magnetic field due to the wire and $A$ is the area of the loop. We can calculate the magnetic field using the Biot-Savart Law:
$$
B = \frac{\mu_0 I}{2\pi r}
$$
where $r$ is the distance from the wire to the center of the loop. At the instant the loop starts moving, $r = 10	ext{ cm}$, so
$$
B = \frac{\mu_0 I}{2\pi(0.1	ext{ m})} = 2.0	imes 10^{-5}	ext{ T}
$$
The area of the loop is $A = (0.1	ext{ m})^2 = 1.0	imes 10^{-2}	ext{ m}^2$. Therefore, the magnitude of the induced emf is
$$
\epsilon = -\frac{d\Phi}{dt} = -B\frac{dA}{dt} = -Bv = -(2.0	imes 10^{-5}	ext{ T})(0.1	ext{ m/s}) = -2.0	imes 10^{-6}	ext{ V}
$$
The direction of the induced current is given by the right-hand rule, which indicates that the current will flow in a clockwise direction when viewed from above. The magnitude of the induced current is given by Ohm's Law:
$$
I = \frac{\epsilon}{R} = \frac{-2.0	imes 10^{-6}	ext{ V}}{0.02	ext{ }\Omega} = \boxed{0.10	ext{ A}}
$$
Note that the question asks for the magnitude of the current, not the charge, so the answer is not given in units of coulombs (C).

The Long Straight Wire in the Figure Has Current I μ\muμ

The Long Straight Wire in the Figure Has Current I $\mu$