A long solenoid with 10.0 text{ turns/cm} and | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The magnetic field due to a long solenoid at any point inside it is directed along its axis: $B_s = \mu_0 n I_s$. Given $n = 10 	ext{ turns/cm} =

Vedclass · Accepted Answer

$3.74$

Vedclass · Answer

(C) The magnetic field due to a long solenoid at any point inside it is directed along its axis: $B_s = \mu_0 n I_s$. Given $n = 10 	ext{ turns/cm} = 1000 	ext{ turns/m}$ and $I_s = 7 	imes 10^{-3} 	ext{ A}$. Thus, $B_s = 4\pi 	imes 10^{-7} 	imes 1000 	imes 7 	imes 10^{-3} = 28\pi 	imes 10^{-7} 	ext{ T}$.The magnetic field due to the straight conductor at a distance $r = 5 	ext{ cm} = 0.05 	ext{ m}$ is directed tangentially: $B_c = \frac{\mu_0 I_c}{2\pi r}$.The resultant magnetic field makes an angle $	heta = 60^{\circ}$ with the axis. Therefore, $	an(60^{\circ}) = \frac{B_c}{B_s}$.$\sqrt{3} = \frac{\mu_0 I_c / (2\pi r)}{\mu_0 n I_s} = \frac{I_c}{2\pi r n I_s}$.$I_c = \sqrt{3} 	imes 2\pi r n I_s = 1.7 	imes 2 	imes 3.14 	imes 0.05 	imes 1000 	imes 7 	imes 10^{-3}$.$I_c = 1.7 	imes 6.28 	imes 0.05 	imes 7 = 3.7366 	ext{ A} \approx 3.74 	ext{ A}$.

Similar Questions

Assertion: The magnetic field produced by a current-carrying solenoid is independent of its length and cross-sectional area.
Reason: The magnetic field inside the solenoid is uniform.

$A$ toroid is:

$A$ toroid has a core (non-ferromagnetic) of inner radius $24 \ cm$ and outer radius $26 \ cm$ around which $2000$ turns of a wire are wound. If the current in the wire is $12 \ A$, the magnetic field inside the core of the toroid is:

Assertion : If the current in a solenoid is reversed in direction while keeping the same magnitude,the magnetic field energy stored in the solenoid decreases.
Reason : Magnetic field energy density is proportional to the square of the current.

Vedclass Test Series

Exam Paper Generator

Online Exam Module