Unit of Magnetic Flux: SI Unit & CGS Units

Ans. Tesla is equal to 1NA^-1m^-1.

Ans. The magnetic field increases stronger when the magnetic field lines are close.

Ans. Magnetic flux density is the number of magnetic field lines crossing unit area that determines the strength of the magnetic field.

Ans. On a charged particle, the magnetic force to electric force ratio is 1.

Ans. Magnetic flux is the number of magnetic field lines passing through the given closed surface. Magnetic flux can be found out by integrating the normal component of magnetic field (B) over the surface. It is commonly represented as φ or φ_B .Weber (Wb) is the SI unit of magnetic flux.

φ_B = B.A = BAcosθ

Where B is the magnetic field, A is the area and θ is the angle at which magnetic field lines pass through the given surface area.

So, the correct answer is option (c).

Ans. 

• Engineers can readily compute the voltage generated by an electric generator using the magnetic flux description, even when the magnetic field is convoluted.
• Although we've only looked at magnetic flux recorded on a flat test-area so far, we can make our test-area any form we like. In reality, a closed surface, such as a sphere, can be used to enclose a region of interest. Because of Gauss's law for magnetism, physicists are particularly interested in closed surfaces. 
• Because magnets always have two poles, there is no way for a magnetic monopole to exist inside a closed surface (as far as we know). As a result, the net magnetic flux across such a closed surface is always zero, and all magnetic field lines entering the closed surface are perfectly balanced by field lines exiting. This information aids in the solution of magnetic field problems.

Ans. The formula for calculating the magnetic flux is given by the formula as follows.

φ = BA

Where B is the magnetic field produced and A is the area of the coil.

The magnetic field produced and the area of the coil equation can be expressed as,

\(\phi=\frac{\mu_{0} i R^{2}}{2\left(R^{2}+x^{2}\right)^{3 / 2}} \times \pi r^{2}\)

Where R is the radius of the one coil, i is the current through the coil and r is the radius of the other coil.

Represent the above equations in terms of the coils P and Q. So, we get,

\(\begin{aligned} \phi_{q} &=\frac{\mu_{0} i R^{2}}{2\left(R^{2}+x^{2}\right)^{3 / 2}} \times \pi r^{2} \\ \phi_{p} &=\frac{\mu_{0} i r^{2}}{2\left(R^{2}+x^{2}\right)^{3 / 2}} \times \pi R^{2} \end{aligned}\)

Divide the above equations in order to simplify this complicated equation.

\(\begin{aligned} &\frac{\phi_{p}}{\phi_{q}}=\frac{2}{3} \frac{\left(R^{2}+x^{2}\right) \frac{3}{2}}{\left(r^{2}+x^{2}\right) \frac{3}{2}} \\ &\Rightarrow \frac{\phi_{p}}{10^{-3}}=\frac{2}{3} \end{aligned}\)

Continue the further calculation to find the value of the flux through the coil P.

Φp = 6.67×10⁻⁴Wb

Therefore, the value of the flux through the coil P when the current of 2 A passes through the coil Q is 6.67×10⁻⁴Wb.

The value of the flux through the coil P is obtained to be equal to 6.67×10⁻⁴Wb. So, the correct answer is “Option b”.

Ans. Magnetic flux is defined as the product of the magnetic field, surface area, and the angle between B and a unit vector perpendicular to the surface with the formula Φm​=BAcosθ. 

Here, the loop is positioned perpendicular to the magnetic field BB namely the angle between BB and a unit vector normal to the surface θ=0^∘. Putting the values in the magnetic flux formula, we

Φm​=BAcosθ

=(0.1)(0.03×0.05)cos0^∘

=15×10⁻⁵Wb

=0.15mWb​

Ans. A metal ring is held horizontally and a bar magnet is dropped through the ring with its length along the axis of the ring. The magnet falls with an acceleration less than g

As the magnet falls, the magnetic flux linked with the ring increases. This induces emf in the ring which opposes the motion of the falling magnet, hence a<g.