The percentage change in \(B\) is given by: \[ \% \text{ change in } B = \frac{B_{\text{new}} - B_{\text{old}}}{B_{\text{old}}} \times 100\% \] Substitute the values: \[ = \frac{\mu_{\text{ni}} - \mu_{\text{ni0}}}{\mu_{\text{ni0}}} \times 100\% = \frac{\mu - \mu_0}{\mu_0} \times 100\% \] \[ = \frac{(\mu_0 \mu_r - \mu_0)}{\mu_0} \times 100\% \] \[ = (\mu_r - 1) \times 100\% \] Thus, the percentage change is: \[ \chi_n \times 100\% = 1.2 \times 10^{-3} \, \% \] \[ \boxed{\text{Percentage change in } B = 1.2 \times 10^{-3} \, \% } \]
Two long parallel wires X and Y, separated by a distance of 6 cm, carry currents of 5 A and 4 A, respectively, in opposite directions as shown in the figure. Magnitude of the resultant magnetic field at point P at a distance of 4 cm from wire Y is \( 3 \times 10^{-5} \) T. The value of \( x \), which represents the distance of point P from wire X, is ______ cm. (Take permeability of free space as \( \mu_0 = 4\pi \times 10^{-7} \) SI units.) 
A particle of charge $ q $, mass $ m $, and kinetic energy $ E $ enters in a magnetic field perpendicular to its velocity and undergoes a circular arc of radius $ r $. Which of the following curves represents the variation of $ r $ with $ E $?
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : If oxygen ion (O\(^{-2}\)) and Hydrogen ion (H\(^{+}\)) enter normal to the magnetic field with equal momentum, then the path of O\(^{-2}\) ion has a smaller curvature than that of H\(^{+}\).
Reason R : A proton with same linear momentum as an electron will form a path of smaller radius of curvature on entering a uniform magnetic field perpendicularly.
In the light of the above statements, choose the correct answer from the options given below
Let one focus of the hyperbola \( H : \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 \) be at \( (\sqrt{10}, 0) \) and the corresponding directrix be \( x = \dfrac{9}{\sqrt{10}} \). If \( e \) and \( l \) respectively are the eccentricity and the length of the latus rectum of \( H \), then \( 9 \left(e^2 + l \right) \) is equal to:
