Magnetic Effects of Electric Current - Compare DC and AC, frequency of AC, and domestic electric circuits

A constant current of $2 \text{ A}$ flows through a straight wire. If the distance from the wire is doubled, what happens to the magnitude of the magnetic field ($B$)?

The magnetic field will become half of its original value.

The magnetic field produced by a straight wire is inversely proportional to the distance from the wire ($B \propto \frac{1}{r}$). If $r$ becomes $2r$, then $B$ becomes $\frac{B}{2}$.

An electron enters a magnetic field ($B$) at a right angle, moving from North to South. The magnetic field is directed from West to East. Determine the direction of the force ($F$) acting on the electron.

The force acts vertically upwards (out of the plane).

Since the electron moves North to South, the conventional current ($I$) is South to North. Using Fleming's Left-Hand Rule: Forefinger (Field) points East, Middle finger (Current) points North. The Thumb then points upwards, indicating the direction of Force ($F$).

Calculate the force acting on a $0.5 \text{ m}$ wire carrying a $DC$ of $4 \text{ A}$ placed perpendicular to a magnetic field of $0.1 \text{ T}$.

$F = 0.1 \times 4 \times 0.5 = 0.2 \text{ N}$

Using the formula $F = BIl \sin \theta$, where $\theta = 90^{\circ}$ (so $\sin 90^{\circ} = 1$), we substitute the values: $F = (0.1 \text{ T})(4 \text{ A})(0.5 \text{ m}) = 0.2 \text{ Newtons}$.

In a standard Indian domestic circuit, the potential difference between the live wire and neutral wire is $220 \text{ V}$ and the frequency is $50 \text{ Hz}$. Calculate the time interval between two consecutive instances where the current becomes zero.

$f = 50 \text{ Hz}$ $T = \frac{1}{f} = \frac{1}{50} \text{ s} = 0.02 \text{ s}$ In one full cycle ($T$), the AC current becomes zero twice. Therefore, the time interval between two consecutive zeros is: t = \frac{T}{2} = \frac{0.01 \text{ s}} $t = \frac{1}{100} \text{ s}$

Alternating current changes direction periodically. For a $50 \text{ Hz}$ supply, there are $100$ such direction changes per second, meaning the current passes through the zero value every $0.01$ seconds.

In a domestic circuit, two appliances, a lamp of resistance $R_1 = 440 \text{ Ω}$ and a heater of resistance $R_2 = 44 \text{ Ω}$, are connected in parallel to an $AC$ source of $220 \text{ V}$. A fuse of rating $5 \text{ A}$ is placed in the live wire. Calculate the total current drawn from the source and determine if the fuse will blow when both appliances are switched on.

$\text{Given: } V = 220 \text{ V}, R_1 = 440 \text{ Ω}, R_2 = 44 \text{ Ω}$ $\text{Current through lamp, } I_1 = \frac{V}{R_1} = \frac{220}{440} = 0.5 \text{ A}$ $\text{Current through heater, } I_2 = \frac{V}{R_2} = \frac{220}{44} = 5 \text{ A}$ $\text{Total current, } I_{total} = I_1 + I_2 = 0.5 + 5 = 5.5 \text{ A}$ $\text{Since } 5.5 \text{ A} > 5 \text{ A} \text{ (fuse rating), the fuse will blow.}$

In a domestic circuit, appliances are connected in parallel so they all receive the same potential difference ($220 \text{ V}$ in India). The total current is the sum of currents through individual branches. If this sum exceeds the safety limit (fuse rating), the fuse wire melts due to Joule heating, breaking the circuit to prevent fire or damage.