A metal rectangular loop (height h and width w) with resistance R is fixed in place with one-third of its length located inside a region of space where there is a time-varying magnetic field B=B0−bt pointing out of the page.a) Determine the magnitude and direction of the current I(t) induced in the loop.b) If the loop were not fixed in place, it would move due to the magnetic force exerted on it by the external magnetic field. What is the magnitude of the magnetic force felt by the loop? What direction would the loop move towards, if it were not fixed in place?c) Determine the torque on the loop about its center of mass due to the magnetic field.
The Correct Answer and Explanation is:
Part (a): Induced Current
We are given a metal rectangular loop of dimensions hhh (height) and www (width), with resistance RRR, placed in a region where there is a time-varying magnetic field B=B0−btB = B_0 – btB=B0−bt pointing out of the page. The length of the loop inside the magnetic field is one-third of its total length.
To find the induced current in the loop, we use Faraday’s Law of Induction, which states that the induced electromotive force (EMF) is equal to the negative rate of change of the magnetic flux through the loop:E=−dΦBdt\mathcal{E} = -\frac{d\Phi_B}{dt}E=−dtdΦB
Where the magnetic flux ΦB\Phi_BΦB through the loop is given by:ΦB=B⋅A\Phi_B = B \cdot AΦB=B⋅A
Here, AAA is the area of the loop inside the magnetic field. Since one-third of the loop is inside the magnetic field, the area A=13⋅h⋅wA = \frac{1}{3} \cdot h \cdot wA=31⋅h⋅w. Given that the magnetic field is varying over time, we write the flux as:ΦB(t)=(B0−bt)⋅(13hw)\Phi_B(t) = (B_0 – bt) \cdot \left(\frac{1}{3} h w\right)ΦB(t)=(B0−bt)⋅(31hw)
Now, the induced EMF is:E(t)=−ddt[(B0−bt)⋅13hw]\mathcal{E}(t) = -\frac{d}{dt} \left[ \left( B_0 – bt \right) \cdot \frac{1}{3} h w \right]E(t)=−dtd[(B0−bt)⋅31hw]
Taking the derivative:E(t)=−13hw⋅(−b)=13bhw\mathcal{E}(t) = – \frac{1}{3} h w \cdot (-b) = \frac{1}{3} b h wE(t)=−31hw⋅(−b)=31bhw
Now, the current induced in the loop is given by Ohm’s law:I(t)=E(t)R=13bhwRI(t) = \frac{\mathcal{E}(t)}{R} = \frac{\frac{1}{3} b h w}{R}I(t)=RE(t)=R31bhw
Thus, the induced current is:I(t)=bhw3RI(t) = \frac{b h w}{3R}I(t)=3Rbhw
The direction of the current follows Lenz’s Law, which states that the induced current will flow in such a way as to oppose the change in magnetic flux. Since the magnetic field is decreasing with time, the current will create its own magnetic field to oppose this decrease. Therefore, the current will flow in a clockwise direction when viewed from above.
Part (b): Magnetic Force
The magnetic force on the loop results from the interaction between the current and the magnetic field. The force on a segment of current-carrying wire in a magnetic field is given by:F=ILBF = I L BF=ILB
Where LLL is the length of the wire in the magnetic field and BBB is the magnetic field strength. The total length of the loop inside the magnetic field is w3\frac{w}{3}3w, and the magnetic field is B=B0−btB = B_0 – btB=B0−bt.
The force on this segment of the wire is:F=I⋅w3⋅B=(bhw3R)⋅w3⋅(B0−bt)F = I \cdot \frac{w}{3} \cdot B = \left( \frac{b h w}{3R} \right) \cdot \frac{w}{3} \cdot (B_0 – bt)F=I⋅3w⋅B=(3Rbhw)⋅3w⋅(B0−bt)
Simplifying:F=bhw29R⋅(B0−bt)F = \frac{b h w^2}{9R} \cdot (B_0 – bt)F=9Rbhw2⋅(B0−bt)
The direction of the force is due to the interaction between the current and the magnetic field. Using the right-hand rule, the force will be directed along the plane of the loop, and since the magnetic field is pointing out of the page, the force on the current-carrying wire will push the loop to the right.
Part (c): Torque
The torque τ\tauτ on the loop about its center of mass is given by:τ=IABsin(θ)\tau = I A B \sin(\theta)τ=IABsin(θ)
Where AAA is the area of the loop, BBB is the magnetic field, and θ\thetaθ is the angle between the magnetic field and the normal to the loop. In this case, since the magnetic field is perpendicular to the plane of the loop, sin(θ)=1\sin(\theta) = 1sin(θ)=1, and the torque becomes:τ=I⋅A⋅B=(bhw3R)⋅(hw3)⋅(B0−bt)\tau = I \cdot A \cdot B = \left( \frac{b h w}{3R} \right) \cdot \left( \frac{h w}{3} \right) \cdot (B_0 – bt)τ=I⋅A⋅B=(3Rbhw)⋅(3hw)⋅(B0−bt)
Simplifying:τ=bh2w29R⋅(B0−bt)\tau = \frac{b h^2 w^2}{9R} \cdot (B_0 – bt)τ=9Rbh2w2⋅(B0−bt)
This torque will cause the loop to rotate about its center of mass. The direction of rotation depends on the sign of bbb. If b>0b > 0b>0, the loop will rotate in the direction that opposes the decreasing magnetic field.