A constant current of 2A is passing through a solenoid of inductance 12H. Find Potential Difference across solenoid.

Collegedunia Team

Content Curator

To find the potential difference across a solenoid, we can use the formula: V = L dI/dt

where V is the potential difference, L is the inductance of the solenoid, and dI/dt is the rate of change of current.

Given, a constant current of 2A is passing through a solenoid of inductance 12H.

Since the current is constant, dI/dt = 0.

Therefore, the potential difference across the solenoid is: V = L dI/dt = 12 x 0 = 0 volts

Hence, the potential difference across the solenoid is 0 volts. This is because there is no rate of change of current, so there is no induced EMF across the solenoid.

Also check:

Important Concepts Related to Magnetic Field
Moving Charges and Magnetism Important Questions	Magnetic Poles And Their Significance	Einstein Field Equation
Curie Weiss Law	Bar Magnet as an Equivalent Solenoid	Magnetic Declination
Magnetic Flux	Electromagnetism	Faraday Constant Value

CBSE CLASS XII Related Questions

1.
A charged particle $+q$ in an electric field $\vec{E}$ experiences a force in the direction of the electric field. As a result, its kinetic energy changes. Similarly, the charged particle also experiences a force when it moves in a magnetic field $\vec{B}$. But this magnetic force is perpendicular to both velocity $\vec{v}$ of the charged particle and the magnetic field $\vec{B}$, so it cannot change the kinetic energy of the charged particle. Consider two charged particles 1 and 2 of masses $m$ and $\frac{m}{2}$ having charges $-q$ and $+2q$ respectively. They are accelerated from rest through the same potential difference $V$ and acquire kinetic energy $K_1$ and $K_2$. Then they enter in a region of uniform magnetic field $\vec{B}$ perpendicular to their velocities.
2.
Read the following paragraph and answer the questions that follow.
In an experiment with convex lens of focal length f, the screen is fixed at a distance D from the object. A student slowly moves the lens away from the object towards the screen and finds that she is able to form sharp image of the object for two positions of the lens. The distance between these two positions of the lens is d.
3.
Capacitors are manufactured with certain standard capacitances and working voltages. However, these standard values may not be the ones that are actually needed in a particular application. Two or more capacitors can be grouped in series or in parallel to achieve desired capacitance and voltage. When connected in series, the total capacitance decreases while the voltage rating increases, whereas in parallel connections, the total capacitance increases and maintains the same voltage rating. A capacitor stores energy in the electric field between its plates and stored energy is proportional to the square of the voltage and capacitance $U = \frac{1}{2}CV^2$, where symbols have their usual meanings.
Two capacitors, one of $3 \ \mu$F and the other of $6 \ \mu$F, are connected in series in the circuit as shown in the figure, for a long time. }
4.
The resistance of a metal wire at $ 20^\circ \text{C} $ is $ 1.05 \, \Omega $ and at $ 100^\circ \text{C} $ is $ 1.38 \, \Omega $. Determine the temperature coefficient of resistivity of this metal.
View Solution
5.
Read the following paragraph and answer the questions that follow.
A p-type or n-type semiconductor can be converted into a p-n junction by doping it with suitable impurity. The motion of majority charge carriers causes diffusion current across the junction while the barrier electric field causes motion of minority carriers for drift current. In case of unbiased diode, the diffusion and drift currents are equal. This equilibrium is disturbed by the biasing batteries. Diodes, therefore, allow currents in one direction. This property of diode is used in making rectifiers.
6.
Two metal spheres of radii $r_1$ and $r_2$ ($> r_1$) having charges $q_1$ and $q_2$ respectively kept in air, are brought in contact. Which of the following statements is not correct ?
- The total charge of the two spheres is conserved.
- Both spheres attain the same potential.
- The final potential of the system equals $\frac{1}{4\pi\epsilon_0} \frac{(q_1 + q_2)}{(r_1 + r_2)}$
- The final potential of the system equals $\frac{1}{4\pi\epsilon_0} \frac{(q_1 + q_2) (r_1 + r_2)}{r_1 r_2}$
View Solution