Question

A solenoid of length $0.4m$, having $500turns$and $3A$current flows through it. A coil of radius $0.01m$ and have $10$ turns and carries current of $0.4A$has to placed such that its axis is perpendicular to the axis of solenoid, then torque on coil will be
$\left( A \right)5.92 \times {10^{ - 7}}Nm$
$\left( B \right)5.92 \times {10^{ - 5}}Nm$
$\left( C \right)5.92 \times {10^{ - 4}}Nm$
$\left( D \right)5.92 \times {10^{ - 3}}Nm$

Answer 1

Electric field is caused by stationary charges and magnetic field are caused by moving electric charges. Ampere’s circuital law states the relationship between the magnetic and the electric field. Here we are considering a long solenoid. Use the Ampere circuit law to obtain the torque.

Formula used:
$B = \dfrac{{{\mu _0}IN}}{l}$
${\mu _0}$ is the permeability of the medium, $i$ is the current enclosed by the closed path and $B$ is the magnetic field.

Complete step by step answer:
Ampere’s circuital law tells us about the relationship between the magnetic and the electric field. This law gives that the integral of (B) in an imaginary closed path is equal to the product of current enclosed by the closed path and permeability of the medium. The force will be zero, when the conductor is parallel to the magnetic field. The force will be maximum, when it is perpendicular to the magnetic field
Number of turns are wound around a cylinder by using an insulated wire called solenoid.
Firstly, we know that a torque is acting on the body
$\tau = MB$
Where $M,B$ are the magnetic moment and magnetic field.
$M = iA$
Where $i,A$ is the current and the area of the solenoid.
Where magnetic field due to solenoid
$B = {\mu _0}ni$
Where n is the number of turns per unit length $n = \dfrac{N}{l}$
$B = \dfrac{{{\mu _0}Ni}}{l}$
Here $i = 3A,l = 0.4m,N = 500$
$B = \dfrac{{4 \times {{10}^{ - 7}} \times 500 \times 3}}{{0.4}}$

Then the torque on coil is given by
$\tau = MB\sin 90$
$\tau = {N_1}{i_1}A\dfrac{{{\mu _0}Ni}}{l}$
Where ${i_1} = 0.4$,${N_1} = 10turns$,$A = \pi {r^2} = \pi \times {\left( {0.01} \right)^2}$
Here ${i_1},{N_1},A$is the current, no of turns and area of the coil.
Substitute the value of B in the torque equation,
$\tau = 10 \times 0.4 \times \pi \times {0.01^2} \times \dfrac{{4\pi \times {{10}^{ - 7}} \times 500 \times 3}}{{0.4}}$
$\tau = 5.92 \times {10^{ - 6}}Nm$

Hence, the correct answer is option (A).

Note: A magnetic field is generated, whenever a current travels through a conductor. The force will be zero, when the conductor is parallel to the magnetic field. The force will be maximum, when it is perpendicular to the magnetic field. The Magnetic field lines around a conductor carrying current are concentric circles whose centres lie on the wire. Right-Hand Thumb Rule determines the direction of magnetic field lines.