Question

In a certain region of space near Earth’s surface, a uniform horizontal magnetic field of magnitude B exists above a level defined to be y = 0. Below y = 0, the field abruptly becomes zero (Fig. 29–63). A vertical square wire loop has resistivity ρ, mass density ρ_m, diameter d, and side length ℓ. It is initially at rest with its lower horizontal side at y = 0 and is then allowed to fall under gravity, with its plane perpendicular to the direction of the magnetic field. (a) While the loop is still partially immersed in the magnetic field (as it falls into the zero-field region), determine the magnetic “drag” force that acts on it at the moment when its speed is υ. (b) Assume that the loop achieves a constant terminal velocity V_T before its upper horizontal side exits the field. Determine a formula for V_T. (c) If the loop is made of copper and B = 0.80 T, find V_T.

Accepted Answer

Step 1: Begin by analyzing the situation described in the problem. The wire loop is falling under gravity through a region with a uniform horizontal magnetic field above y = 0, and no magnetic field below y = 0. The loop experiences a magnetic drag force due to the induced current caused by its motion through the magnetic field. This force opposes the motion of the loop. Step 2: To determine the magnetic drag force (part a), use Faraday's law of induction to calculate the induced electromotive force (EMF) in the loop. The EMF is given by $$ \mathcal{E} = - \frac{d\Phi_B}{dt} $$, where $$ \Phi_B  is $$the magnetic flux. The flux is $$ \Phi_B = B \cdot A $$, where $$ A  is $$the area of the loop in the magnetic field. As the loop falls, the area exposed to the field changes, leading to an EMF. Step 3: The induced current $$ I  in $$the loop can be calculated using Ohm's law: $$ I = \frac{\mathcal{E}}{R} $$, where $$ R  is $$the resistance of the loop. The resistance is determined by the resistivity $$ \rho $$, the length of the wire $$ L $$, and its cross-sectional area $$ A_c $$: $$ R = \frac{\rho L}{A_c} . $$Substitute the values for $$ L  ($$the total length of the square loop) and $$ A_c  ($$based on the wire's diameter $$ d ). $$Step 4: The magnetic drag force $$ F_{drag}  is $$given by $$ F_{drag} = I \cdot \ell \cdot B $$, where $$ \ell  is $$the length of the side of the loop in the magnetic field. Substitute the expression for $$ I $$ from Step 3 into this formula to find $$ F_{drag}  as a $$function of the loop's velocity $$ \upsilon . $$Step 5: For part (b), at terminal velocity $$ V_T $$, the magnetic drag force balances the gravitational force acting on the loop. Set $$ F_{drag} = F_{gravity} $$, where $$ F_{gravity} = P_m \cdot g \cdot \ell^2  ($$mass density times gravitational acceleration times the area of the loop). Solve for $$ V_T $$ using the expressions derived for $$ F_{drag} $$ and $$ F_{gravity} . $$For part (c), substitute the given values for copper's resistivity, $$ B $$, and other parameters into the formula for $$ V_T  to $$find its numerical value.

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Concetti chiave

Magnetic Induction

Lorentz Force

Terminal Velocity