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Question

(D) The magnetic field $B$ inside a long solenoid is given by the formula:$B = \mu_0 n I$where $n$ is the number of turns per unit length and $I$ is t

Vedclass · Accepted Answer

$\frac{2B}{3}$

Vedclass · Answer

(D) The magnetic field $B$ inside a long solenoid is given by the formula:$B = \mu_0 n I$where $n$ is the number of turns per unit length and $I$ is the current flowing through the solenoid.Given that the initial magnetic field is $B = \mu_0 n I$.According to the problem,the new number of turns per unit length $n' = 2n$ and the new current $I' = \frac{1}{3}I$.The new magnetic field $B'$ is given by:$B' = \mu_0 n' I'$$B' = \mu_0 (2n) \left(\frac{1}{3}I\right)$$B' = \frac{2}{3} (\mu_0 n I)$$B' = \frac{2}{3} B$Therefore,the new value of the magnetic field is $\frac{2B}{3}$.

$A$ long solenoid carrying a current produces a magnetic field $B$ along its axis. If the number of turns per $cm$ is doubled and the current is made $\left(\frac{1}{3}\right)^{rd}$ of its original value,then the new value of the magnetic field will be:

Similar Questions

$A$ long straight wire of a circular cross-section with radius $a$ carries a steady current $I$. The current $I$ is uniformly distributed across this cross-section. The plot of the magnitude of the magnetic field $B$ with distance $r$ from the centre of the wire is given by:

The magnetic field at point $P$ in the given figure, due to a current $I$ flowing through a cylindrical conductor of radius $R$, is:

$A$ long solenoid carrying current $I_1$ produces a magnetic field $B_1$ along its axis. If the current is reduced to $20 \%$ and the number of turns per $cm$ is increased five times,then the new magnetic field $B_2$ is equal to:

The magnetic induction due to an ideal solenoid is independent of

Statement-$1$: Ampere's law can be used to find the magnetic field due to a finite length of a straight current-carrying wire.
Statement-$2$: The magnetic field due to a finite length of a straight current-carrying wire is symmetric about the wire.

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