Electric Current

A positive charge is moving towards south in a space where magnetic field is pointing in the north direction. The moving charge will experience :

[amp_mcq option1=”a deflecting force towards north direction.” option2=”a deflecting force towards east direction.” option3=”a deflecting force towards west direction.” option4=”no deflecting force.” correct=”option4″]

The force experienced by a positive charge (q > 0) moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ is given by the Lorentz force formula: $\vec{F} = q (\vec{v} \times \vec{B})$.

The direction of the velocity $\vec{v}$ is towards south, and the direction of the magnetic field $\vec{B}$ is towards north. These two directions are opposite to each other. The angle between $\vec{v}$ and $\vec{B}$ is 180 degrees.

The magnitude of the cross product $\vec{v} \times \vec{B}$ is $|\vec{v}| |\vec{B}| \sin(\theta)$, where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$. Since $\theta = 180^{\circ}$, $\sin(180^{\circ}) = 0$. Therefore, the magnitude of the force $|\vec{F}| = q |\vec{v}| |\vec{B}| \sin(180^{\circ}) = 0$. The moving charge will experience no deflecting force.

Two resistors $R_1$ and $R_2$ arranged in parallel combination in an electrical closed circuit are made of the same material and of same thickness. If the length of $R_2$ is twice the length of $R_1$, then the total resistance $R$ satisfies

[amp_mcq option1=”$3R = 2R_1$” option2=”$3R = 2R_2$” option3=”$2R = 3R_1$” option4=”$2R = 3R_2$” correct=”option1″]

The correct option is A.

For resistors in parallel, the total resistance $R$ is given by $1/R = 1/R_1 + 1/R_2$. The resistance of a wire is $R = \rho L/A$, where $\rho$ is resistivity, $L$ is length, and $A$ is cross-sectional area.

Given that $R_1$ and $R_2$ are made of the same material ($\rho_1 = \rho_2 = \rho$) and have the same thickness (assuming same cross-sectional area $A_1 = A_2 = A$).
Let the length of $R_1$ be $L_1$. Then $R_1 = \rho L_1 / A$.
The length of $R_2$ is $L_2 = 2L_1$. Then $R_2 = \rho L_2 / A = \rho (2L_1) / A = 2 (\rho L_1 / A) = 2R_1$.
Now, the resistors are in parallel, so $1/R = 1/R_1 + 1/R_2$.
Substitute $R_2 = 2R_1$:
$1/R = 1/R_1 + 1/(2R_1)$
$1/R = (2 + 1) / (2R_1)$
$1/R = 3 / (2R_1)$
$R = (2R_1) / 3$
Multiplying both sides by 3 gives $3R = 2R_1$.

According to Fleming’s right-hand rule, if the forefinger indicates the direction of magnetic field and thumb shows the direction of motion of conductor, then the stretched middle finger will predict the direction of

[amp_mcq option1=”force acting on the conductor” option2=”electric field” option3=”induced current” option4=”current” correct=”option3″]

The correct option is C.

Fleming’s Right-Hand Rule is used to determine the direction of the induced current in a conductor moving in a magnetic field.

According to Fleming’s Right-Hand Rule, if the thumb points in the direction of the motion of the conductor, the forefinger points in the direction of the magnetic field, then the middle finger points in the direction of the induced electric current. This rule is fundamental to understanding the operation of generators. Fleming’s Left-Hand Rule, in contrast, determines the direction of the force on a current-carrying conductor placed in a magnetic field.

The magnetic field produced by a current-carrying straight wire at a point outside the wire depends

[amp_mcq option1=”inversely on the distance from it” option2=”directly on the distance from it” option3=”inversely at short distances and directly at large distances from it” option4=”directly on the distance (at short distances) and inversely on the distance (at long distances) from it” correct=”option1″]

The magnetic field produced by a current-carrying straight wire at a point outside the wire depends inversely on the distance from it.

The magnitude of the magnetic field ($B$) at a distance ($r$) from a long, straight conductor carrying current ($I$) is given by the formula $B = \frac{\mu_0 I}{2\pi r}$, where $\mu_0$ is the permeability of free space.

From the formula $B = \frac{\mu_0 I}{2\pi r}$, it is clear that for a constant current $I$, the magnetic field strength $B$ is inversely proportional to the distance $r$ from the wire ($B \propto \frac{1}{r}$). This relationship holds true for points outside the wire.

What is the current required to light a 60 W incandescent bulb in a domestic supply of 240 V ?

[amp_mcq option1=”0·5 A” option2=”0·25 A” option3=”1·0 A” option4=”5·0 A” correct=”option2″]

The current required is 0.25 A.

The power ($P$) consumed by an electrical device is related to the voltage ($V$) across it and the current ($I$) flowing through it by the formula $P = V \times I$.

Given Power $P = 60$ W and Voltage $V = 240$ V. Using the formula $P = V \times I$, we can find the current $I$: $I = \frac{P}{V} = \frac{60 \text{ W}}{240 \text{ V}} = \frac{60}{240} \text{ A} = \frac{1}{4} \text{ A} = 0.25 \text{ A}$.

The frequency of an alternating current is 3 Hz. It implies that

[amp_mcq option1=”there are 6 cycles/s” option2=”there are 3 cycles/s” option3=”there are 2 cycles/s” option4=”there is only 1 cycle/s” correct=”option2″]

A frequency of 3 Hz implies that there are 3 cycles per second.

Frequency is defined as the number of cycles or oscillations per unit time. The SI unit of frequency is Hertz (Hz), where 1 Hz equals 1 cycle per second.

In the context of alternating current (AC), one cycle represents a complete sequence of changes in voltage or current from zero to maximum positive, back to zero, to maximum negative, and back to zero again. A frequency of 3 Hz means this complete cycle repeats 3 times every second.

The electric field lines from an isolated positively charged conducting sphere are

[amp_mcq option1=”tangential to the conducting surface” option2=”at right angles to the conducting surface and towards the centre of the sphere” option3=”at any angle to the conducting surface” option4=”at right angles to the conducting surface and outwards from the centre of the sphere” correct=”option4″]

For a conductor in electrostatic equilibrium, the electric field lines are always perpendicular to the surface, and for a positively charged object, they point outwards.

In electrostatic equilibrium, the electric field inside a conductor is zero, and any net charge resides on the surface. If the electric field had a component parallel to the surface, charges would move along the surface, and the conductor would not be in equilibrium. Therefore, the electric field lines must be perpendicular (at right angles) to the surface of the conductor. For a positively charged object, electric field lines originate from the positive charges and point away from them. For a sphere, this direction is radially outwards from the centre.

For a negatively charged conducting sphere, the electric field lines would also be at right angles to the surface but would point inwards towards the centre of the sphere, terminating on the negative charges on the surface.

An electric wire of resistance 50 ohm is cut into five equal wires. These wires are then connected in parallel. What is the equivalent resistance of this combination?

[amp_mcq option1=”2 ohm” option2=”10 ohm” option3=”0.5 ohm” option4=”5 ohm” correct=”option1″]

When a wire of resistance 50 ohm is cut into five equal parts, each part will have one-fifth of the original resistance. Connecting these five equal resistances in parallel results in a combined resistance calculated using the parallel resistance formula.

The resistance of a uniform wire is directly proportional to its length. So, if the wire is cut into five equal parts, the resistance of each part is R’ = R/5 = 50 ohm / 5 = 10 ohm. When n equal resistors (each of resistance r) are connected in parallel, the equivalent resistance (Req) is given by Req = r/n. In this case, r = 10 ohm and n = 5. Thus, Req = 10 ohm / 5 = 2 ohm.

Alternatively, the formula for resistors in parallel can be used: 1/Req = 1/R1 + 1/R2 + … + 1/Rn. For five equal resistances of 10 ohm in parallel: 1/Req = 1/10 + 1/10 + 1/10 + 1/10 + 1/10 = 5/10 = 1/2. Therefore, Req = 2 ohm. Connecting resistors in parallel decreases the overall resistance.

Which of the following statements correctly explains/explain the existence of a positive force between two electric charges?

• Both the charges are positive.
• Both the charges are negative.
• Both the charges are oppositely charged.

Select the correct answer using the code given below.

[amp_mcq option1=”1 only” option2=”2 only” option3=”1 and 2 only” option4=”1, 2 and 3″ correct=”option3″]

A positive force between two electric charges indicates a repulsive force. Repulsive forces exist between like charges, meaning both charges are either positive or both are negative.

According to Coulomb’s Law, the force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. The force is repulsive if the charges are of the same sign (both positive or both negative) and attractive if they are of opposite signs (one positive and one negative). A positive force value in physics convention often represents repulsion, while a negative force value represents attraction.

Statement 1 describes the case of two positive charges, which repel each other. Statement 2 describes the case of two negative charges, which also repel each other. Statement 3 describes the case of oppositely charged particles (e.g., positive and negative), which attract each other, resulting in an attractive (negative) force. Therefore, only statements 1 and 2 correctly explain the existence of a positive (repulsive) force.

Which one of the following formulas does *not* represent electrical power?

[amp_mcq option1=”I²R” option2=”IR²” option3=”VI” option4=”V²/R” correct=”option2″]

The formula that does *not* represent electrical power is B) IR².

– Electrical power (P) is the rate at which electrical energy is transferred or consumed.
– The basic formulas for power are P = VI, where V is voltage and I is current.
– Using Ohm’s Law (V = IR), we can derive other forms:
– Substituting V = IR into P = VI gives P = (IR) * I = I²R. (Option A)
– Substituting I = V/R into P = VI gives P = V * (V/R) = V²/R. (Option D)
– Option C, VI, is also a standard formula for power.

– The formula IR² is incorrect. The power dissipated by a resistor is I²R.
– The unit of power is the Watt (W).