b) Figure shows a rectangular metal frame ABCD placed in a uniform magnetic field of flux density 4.5 10^ 3\ T . [figure in the original paper] i) Calculate the force experienced by side AB. [2] ii)…

Answer to Question b)

i) Force experienced by side AB

The force on a current-carrying conductor in a magnetic field is given by: $$F=BIL$$ where:

• $B=4.5\times 10^{-3}\mathrm{T}$ (magnetic flux density),
• $I$ = current through the conductor,
• $L$ = length of the conductor.

Assumption: The figure shows a rectangular frame ABCD with side AB of length $L=0.1\mathrm{m}$ (a typical value for such problems; if the figure provides a different length, replace it). The current $I$ flows through AB.

Since the problem does not specify the current, we assume the frame is connected to a battery or current source, and the current $I$ is given or implied. If the current is not provided, we cannot compute the force numerically. However, if we assume $I=2\mathrm{A}$ (a common value in such problems), then:

$$F=(4.5\times 10^{-3})\times 2\times 0.1=9\times 10^{-4}\mathrm{N}$$

If the current is not given, the answer should be left in terms of $I$: $$F=BIL=(4.5\times 10^{-3})\times I\times L$$

Final Answer (if $I=2\mathrm{A}$ and $L=0.1\mathrm{m}$): The force experienced by side AB is $9\times 10^{-4}\mathrm{N}$.

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ii) Why side BC does not experience force?

The force on a current-carrying conductor in a magnetic field is given by: $$\vec{F}=I(\vec{L}\times \vec{B})$$ where:

• $\vec{L}$ is the length vector of the conductor,
• $\vec{B}$ is the magnetic field vector.

For side BC to experience a force, the angle between $\vec{L}$ and $\vec{B}$ must be non-zero (i.e., they must not be parallel or antiparallel). If the magnetic field is uniform and perpendicular to the plane of the frame, then:

• Side AB is parallel to the magnetic field (or at 0°), so $\vec{L}\times \vec{B}=0$ (no force).
• Side BC is perpendicular to the magnetic field (or at 90°), so it experiences maximum force.

However, if the magnetic field is parallel to the plane of the frame (i.e., in the same plane as ABCD), then:

• Side AB is perpendicular to the magnetic field (or at 90°), so it experiences force.
• Side BC is parallel to the magnetic field (or at 0°), so $\vec{L}\times \vec{B}=0$ (no force).

Conclusion: Side BC does not experience force because it is parallel to the magnetic field, making the angle between $\vec{L}$ and $\vec{B}$ equal to 0°, resulting in zero cross product and hence zero force.

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Answer to Question a) OR

a) Define impedance of LCR circuit.

Impedance ($Z$) of an LCR circuit is the total opposition offered by the circuit to the flow of alternating current. It is a complex quantity given by: $$Z=\sqrt{R^2+(X_L-X_C{)}^2}$$ where:

• $R$ = resistance,
• $X_L=\omega L$ = inductive reactance,
• $X_C=\frac{1}{\omega C}$ = capacitive reactance,
• $\omega$ = angular frequency.

Impedance has magnitude and phase, and its unit is ohm ($\Omega$).

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b) i) Calculate the impedance of the circuit.

Given:

• $L=60\mathrm{mH}=60\times 10^{-3}\mathrm{H}$,
• $C=0.50\mu \mathrm{F}=0.50\times 10^{-6}\mathrm{F}$,
• $R=300\Omega$,
• Assume the frequency $f$ is given (if not, we cannot compute numerically). A typical value for such problems is $f=50\mathrm{Hz}$ (if not provided, state it).

First, compute angular frequency: $$\omega =2\pi f=2\pi \times 50=100\pi rad/s$$

Now, compute $X_L$ and $X_C$: $$X_L=\omega L=100\pi \times 60\times 10^{-3}=6\pi \Omega \approx 18.85\Omega$$ $$X_C=\frac{1}{\omega C}=\frac{1}{100\pi \times 0.50\times 10^{-6}}=\frac{2\times 10^4}{\pi }\Omega \approx 6366.2\Omega$$

Now, compute impedance: $$Z=\sqrt{R^2+(X_L-X_C{)}^2}=\sqrt{300^2+(18.85-6366.2{)}^2}$$ $$Z=\sqrt{90000+(-6347.35{)}^2}=\sqrt{90000+40295600}=\sqrt{40385600}\approx 6355.2\Omega$$

Final Answer: The impedance of the circuit is $6355.2\Omega$ (approximately).

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ii) How does resonance occur in the above LCR circuit? Write the required expression.

Resonance in an LCR circuit occurs when the inductive reactance ($X_L$) equals the capacitive reactance ($X_C$), i.e., $$X_L=X_C$$ At resonance: $$\omega L=\frac{1}{\omega C}$$ $${\omega }^2=\frac{1}{LC}$$ $$\omega =\frac{1}{\sqrt{LC}}$$ The resonant frequency ($f_0$) is: $$f_0=\frac{\omega }{2\pi }=\frac{1}{2\pi \sqrt{LC}}$$

At resonance:

• The impedance is minimum ($Z=R$),
• The current is maximum,
• The phase difference between voltage and current is zero.

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c) i) Identify device X and write the expression for its reactance.

Given: $$V=V_0\sin \omega t$$ $$I=I_0\sin \left(\omega t+\frac{\pi }{2}\right)$$ The current leads the voltage by $\frac{\pi }{2}$ (90°), which is characteristic of a capacitor.

For a capacitor: $$I=C\frac{dV}{dt}$$ The capacitive reactance ($X_C$) is: $$X_C=\frac{1}{\omega C}$$

Final Answer: Device X is a capacitor, and its reactance is: $$X_C=\frac{1}{\omega C}$$

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c) ii) Graphs of voltage and current for X (capacitor)