What is the magnetic field at point P in the | vedclass

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

Question

(A) The magnetic field on the axis of a finite solenoid at a point $P$ is given by the formula:$B = \frac{{{\mu _0}ni}}{2}(\sin 	heta_1 + \sin 	heta

Vedclass · Accepted Answer

$\frac{{{\mu _0}ni}}{4}(\sqrt 3 + 1)$

Vedclass · Answer

(A) The magnetic field on the axis of a finite solenoid at a point $P$ is given by the formula:$B = \frac{{{\mu _0}ni}}{2}(\sin 	heta_1 + \sin 	heta_2)$where $	heta_1$ and $	heta_2$ are the angles subtended by the ends of the solenoid at point $P$.From the figure,$	heta_1 = 60^\circ$ and $	heta_2 = 30^\circ$.Substituting these values into the formula:$B = \frac{{{\mu _0}ni}}{2}(\sin 60^\circ + \sin 30^\circ)$$B = \frac{{{\mu _0}ni}}{2}(\frac{\sqrt{3}}{2} + \frac{1}{2})$$B = \frac{{{\mu _0}ni}}{4}(\sqrt{3} + 1)$

What is the magnetic field at point $P$ in the given figure?

Similar Questions

The magnetic energy stored in a long solenoid of area of cross-section $A$ in a small region of length $L$ is

An infinitely long hollow conducting cylinder with inner radius $R/2$ and outer radius $R$ carries a uniform current density $J$ along its length. The magnitude of the magnetic field,$|\vec{B}|$ as a function of the radial distance $r$ from the axis is best represented by:

If a copper rod carries a direct current,the magnetic field associated with the current will be

State Ampere's circuital law and explain its mathematical expression.

Vedclass Test Series

Exam Paper Generator

Online Exam Module