Question

A metal detector uses a changing magnetic field to detect metallic objects. Suppose a metal detector that generates a uniform magnetic field perpendicular to its surface is held stationary at an angle of 15.0∘ to the ground, while just below the surface there lies a silver bracelet consisting of 6 circular loops of radius 5.00 cm with the plane of the loops parallel to the ground. If the magnetic field increases at a constant rate of 0.0250 T/s, what is the induced emf E? Take the magnetic flux through an area to be positive when B⃗ crosses the area from top to bottom.

Answer 1

Answer:

$-1.14 \cdot 10^{-3} V$

Explanation:

The induced emf in the loop is given by Faraday's Newmann Lenz law:

$\epsilon = - \frac{d \Phi}{dt}$ (1)

where

$d\Phi$ is the variation of magnetic flux

$dt$ is the variation of time

The magnetic flux through the coil is given by

$\Phi = NBA cos \theta$ (2)

where

N = 6 is the number of loops

A is the area of each loop

B is the magnetic field strength

$\theta =15^{\circ}$ is the angle between the direction of the magnetic field and the normal to the area of the coil

Since the radius of each loop is r = 5.00 cm = 0.05 m, the area is

$A=\pi r^2 = \pi (0.05 m)^2=0.0079 m^2$

Substituting (2) into (1), we find

$\epsilon = - \frac{d (NBA cos \theta)}{dt}= -(NAcos \theta) \frac{dB}{dt}$

where

$\frac{dB}{dt}=0.0250 T/s$ is the rate of variation of the magnetic field

Substituting numbers into the last formula, we find

$\epsilon = -(6)(0.0079 m^2)(cos 15^{\circ})(0.0250 T/s)=-1.14 \cdot 10^{-3} V$

Answer 2

Answer:

Induced emf, $\epsilon=-1.13\times 10^{-3}\ volts$

Explanation:

It is given that,

Number of circular loop, N = 6

A uniform magnetic field perpendicular to its surface is held stationary at an angle of 15 degrees to the ground.

Radius of the loop, r = 5 cm = 0.05 m

Change in magnetic field, $\dfrac{dB}{dt}=0.025\ T/s$

Due to the change in magnetic field, an emf will be induced. Let E is the induced emf in the coil. it is given by :

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

$\phi$ = magnetic flux

$\epsilon=\dfrac{d(NBA\ cos\theta)}{dt}$

$\epsilon=-NA\dfrac{d(B)}{dt}$

$\epsilon=6\times \pi (0.05)^2\times 0.025\times cos(15)$

$\epsilon=-1.13\times 10^{-3}\ volts$

So, the induced emf in the loop is $1.13\times 10^{-3}\ volts$ . Hence, this is the required solution.