Question

26. A square loop whose sides are 6.0-cm long is made with copper wire of radius 1.0 mm. If a magnetic field perpendicular to the loop is changing at a rate of 5.0 mT/s, what is the current in the loop?

Answer 1

Answer:

Explanation:

The formula for determining the Emf induced in a loop is:

$\varepsilon = \dfrac{d \phi}{dt}$

$\varepsilon = \dfrac{d (B*A)}{dt}$

$\varepsilon = A \times \dfrac{dB}{dt}$

$\varepsilon = (side (l))^2 \times \dfrac{dB}{dt}$

where;

square area A = ( l²)

l² = 6.0 cm = 6.0 × 10⁻²

∴

$\varepsilon = ( 6.0 \times 10^{-2})^2 \times 5.0 \times 10^{-3} \ T/S$

$\varepsilon =18 \times 10^{6} \ V$

Recall that:

The resistivity of copper = $1.68 \times 10^{-8}$ ohm m

We can as well say that the length of the copper wire = perimeter of the square loop;

The perimeter of the square loop = 4L

Thus, the length of the copper wire  = 4 (6.0 × 10⁻² )m

= 24× 10⁻² m

Finally, the current in the loop is determined from the formula:

V = IR

where,

V = voltage

I = current and R = resistance of the wire

Making "I" the subject:

I = V/R

where;

$R = \dfrac{\rho \times l}{A}$

$R = \dfrac{\rho \times l}{\pi * r^2}$

$R = \dfrac{1.68 *10^{-8} \times 24*10^{-2}}{\pi * (1*10^{-3})^2}$

$R = 0.001283 \ ohms$

∴

$I = \dfrac{18*10^{-6}}{0.001283}$

I = 14.029 mA