Question

At what rate a single conductor should cut the magnetic flux so that current of $1.5\ \text{mA}$ flows through it when a resistance of $5\ \Omega$ is connected across its ends?

• A. $6 \times 10^{-3}\ \text{wb s}^{-1}$
• B. $8 \times 10^{-3}\ \text{wb s}^{-1}$
• C. $4 \times 10^{-4}\ \text{wb s}^{-1}$
• D. $7.5 \times 10^{-3}\ \text{wb s}^{-1}$

Hint:

Remember that $1\ \text{Volt}$ is equivalent to $1\ \text{Weber per second}$ ($\text{Wb s}^{-1}$). Finding the rate of flux cutting is physically identical to calculating the target open-circuit voltage drop across the system using a direct Ohm's Law calculation ($I \times R$).

Answer 1

The Correct Option is D

Step 1: Link the current to the e.m.f.
By Ohm's law, the e.m.f. needed to push a current $I$ through resistance $R$ is $$e = IR.$$
Step 2: Link the e.m.f. to the flux rate.
By Faraday's law, this same e.m.f. equals the rate at which the conductor cuts magnetic flux: $$e = \frac{d\phi}{dt}.$$
Step 3: Combine the two relations.
Setting them equal gives $$\frac{d\phi}{dt} = IR.$$
Step 4: Convert units.
The current is $I = 1.5\ \text{mA} = 1.5\times 10^{-3}\ \text{A}$, and $R = 5\ \Omega.$
Step 5: Substitute the values.
$$\frac{d\phi}{dt} = (1.5\times 10^{-3})\times 5.$$
Step 6: Compute the rate.
$$\frac{d\phi}{dt} = 7.5\times 10^{-3}\ \text{Wb s}^{-1}.$$
\[ \boxed{\frac{d\phi}{dt} = 7.5\times 10^{-3}\ \text{Wb s}^{-1}} \]