A rectangular loop of size 5 cm × 8 cm is lying in x-y plane in a uniform magnetic field $\vec{B} = (2.0 \text{ T}) \hat{k}$. The total magnetic flux linked with the loop is:

Question

A coil has $N$ turns and current passes through it is $I$ ampere then we obtain $L$ of self inductance. Now if current is made doubled then new self inductance be ______ H.

Answer 1

To find the total magnetic flux linked with the loop, we need to use the formula for magnetic flux, which is given by:

$\Phi = \vec{B} \cdot \vec{A}$

Here, $\Phi$ is the magnetic flux, $\vec{B}$ is the magnetic field, and $\vec{A}$ is the area vector. The area vector's magnitude is equal to the area of the loop, and its direction is perpendicular to the plane of the loop.

Given:

Magnetic field, $\vec{B} = (2.0 \, \text{T}) \hat{k}$
Dimensions of the rectangular loop: 5 cm × 8 cm

The loop is in the x-y plane, therefore, the area vector $\vec{A}$ is along the z-axis (since it is perpendicular to the x-y plane). Hence, we have:

$\vec{A} = (A) \hat{k}$

The area of the loop, $A$, is:

$A = \text{length} \times \text{width} = (5 \, \text{cm}) \times (8 \, \text{cm})$

Converting cm to m (since 1 m = 100 cm):

$A = (0.05 \, \text{m}) \times (0.08 \, \text{m})$

Calculating the area:

$A = 0.004 \, \text{m}^2$

Thus, the area vector is:

$ \vec{A} = (0.004 \, \text{m}^2) \hat{k}$

The magnetic flux through the loop is then:

$\Phi = \vec{B} \cdot \vec{A} = (2.0 \, \text{T}) \hat{k} \cdot (0.004 \, \text{m}^2) \hat{k}$

Since both vectors are parallel:

$\Phi = 2.0 \, \text{T} \times 0.004 \, \text{m}^2 = 0.008 \, \text{Wb}$

Hence, the total magnetic flux linked with the loop is:

$\Phi = 8 \times 10^{-3} \, \text{Wb}$

The correct answer is: $8 \times 10^{-3}$ Wb

Answer 2