The magnetic field intensity due to a solenoid at end point $P$ is :

μ_{0} n I

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{μ_{0} n I}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{μ_{0} n I}{2} (cos θ - cos ϕ)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{μ_{0} n I}{2} (sin θ - sin ϕ)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{μ_{0} n I}{2}$
Magnetic field due to solenoid = $\frac{μ_{0} n i (cos θ - cos ϕ)}{2}$
$| B |$ due to solenoid at P = $\frac{μ_{0} n i (cos 0^{o} - cos 90^{o})}{2} = \frac{μ_{0} n i}{2}$

Suggest Corrections

Similar questions

Q. A circular coil of closely wound N turns and radius r carries a current I. If the expressions of The magnetic field at its centre is

\frac{μ_{0} N i}{x R}

. Find x.

Q. The magnetic field inside a tightly wound, long solenoid is B = µ₀ ni. It suggests that the field does not depend on the total length of the solenoid, and hence if we add more loops at the ends of a solenoid the field should not increase. Explain qualitatively why the extra-added loops do not have a considerable effect on the field inside the solenoid.

Q. A tightly-wound solenoid of radius a and length l has n turns per unit length. It carries an electric current i. Consider a length dx of the solenoid at a distance x from one end. This contains n dx turns and may be approximated as a circular current i n dx. (a) Write the magnetic field at the centre of the solenoid due to this circular current. Integrate this expression under proper limits to find the magnetic field at the centre of the solenoid. (b) verify that if l >> a, the field tends to B = µ₀ni and if a >> l, the field tends to

B = \frac{μ_{0} n i l}{2 a}

. Interpret these results.

Q. Derive expression for the intensity of magnetic field at a point in end on position due to a short bar magnet.

From above fromula we can say, magnetic field due to a current carrying toroidal solenoid is independent of its radius.The magnetic field at the ends of a solenoid is

Join BYJU'S Learning Program

The magnetic field intensity due to a solenoid at end point P is :

The correct option is B μ0nI2Magnetic field due to solenoid =μ0ni(cosθ−cosϕ)2|B| due to solenoid at P =μ0ni(cos0o−cos90o)2=μ0ni2

The magnetic field intensity due to a solenoid at end point $P$ is :

The correct option is B $\frac{μ_{0} n I}{2}$
Magnetic field due to solenoid = $\frac{μ_{0} n i (cos θ - cos ϕ)}{2}$
$| B |$ due to solenoid at P = $\frac{μ_{0} n i (cos 0^{o} - cos 90^{o})}{2} = \frac{μ_{0} n i}{2}$