
A metal wheel of radius $10cm$ rotates with its plane at right angles to a magnetic field of induction $10Wb/{m^2}$ at the rate of $5$ $r.p.s$. the $e.m.f$ induced between the end of a spoke will be
A). $2.57V$
B). $1.57V$
C). $0.57V$
D). $0.057V$
Answer
233.1k+ views
Hint: We understand the length and magnetic field rate here so we calculate the induced emf when defined as the angular displacement change rate, which specifies the angular velocity, so we calculate the angular velocity and then we find the induced emf.
Formula used:
Induced emf,
$E = \dfrac{1}{2}Bw{L^2}$
Where,
$w = $ angular velocity of the rod
$B = $ external magnetic field
$L = $ length of the rod
Angular velocity is the ratio of angular displacement to time,
$w = \dfrac{{2\pi n}}{t}$
Where,
$n$ change in angular rotation
$t$ change in time.
Complete step-by-step solution:
Given by,
Length of the rod $L = 10cm$
Angular velocity of rod $w = 5r.p.s$
Induced $e.m.f$
Because of the changes in the magnetic flux through it, it can be described as the generation of a potential difference in a coil. In simpler terms, when the flux connected to a conductor or coil increases, the electromotive force or EMF is said to be induced.
$E = \dfrac{1}{2}Bw{L^2}$
Substituting the given value,
$w = 5 \times 2\pi $$rad/s$
Or
Angular velocity is a vector quantity and is defined as the angular displacement rate of change that specifies an object's angular velocity or rotational velocity and the axis around which the object rotates.
$w = 10\pi rad/s$
Here,
The external magnitude,
$B = 10wb/{m^2}$
We substituting the given value in the above equation,
We get,
$E = \dfrac{1}{2}Bw{L^2}$
Now,
$E = \dfrac{1}{2} \times 10 \times 10\pi \times {0.1^2}$
On simplifying,
$\dfrac{\pi }{2}Volts$
On solving this equation,
We get,
$1.57V$
Hence,
the $e.m.f$ induced between the end of a spoke will be $1.57V$
Thus, option B is the correct answer
Note: A vector field that describes the magnetic effect on the moving electrical charges and magnetic materials as described above according to the induced emf and angular velocity. In a magnetic field, a charge that travels experiences a force perpendicular to its own velocity and to the magnetic field.
Formula used:
Induced emf,
$E = \dfrac{1}{2}Bw{L^2}$
Where,
$w = $ angular velocity of the rod
$B = $ external magnetic field
$L = $ length of the rod
Angular velocity is the ratio of angular displacement to time,
$w = \dfrac{{2\pi n}}{t}$
Where,
$n$ change in angular rotation
$t$ change in time.
Complete step-by-step solution:
Given by,
Length of the rod $L = 10cm$
Angular velocity of rod $w = 5r.p.s$
Induced $e.m.f$
Because of the changes in the magnetic flux through it, it can be described as the generation of a potential difference in a coil. In simpler terms, when the flux connected to a conductor or coil increases, the electromotive force or EMF is said to be induced.
$E = \dfrac{1}{2}Bw{L^2}$
Substituting the given value,
$w = 5 \times 2\pi $$rad/s$
Or
Angular velocity is a vector quantity and is defined as the angular displacement rate of change that specifies an object's angular velocity or rotational velocity and the axis around which the object rotates.
$w = 10\pi rad/s$
Here,
The external magnitude,
$B = 10wb/{m^2}$
We substituting the given value in the above equation,
We get,
$E = \dfrac{1}{2}Bw{L^2}$
Now,
$E = \dfrac{1}{2} \times 10 \times 10\pi \times {0.1^2}$
On simplifying,
$\dfrac{\pi }{2}Volts$
On solving this equation,
We get,
$1.57V$
Hence,
the $e.m.f$ induced between the end of a spoke will be $1.57V$
Thus, option B is the correct answer
Note: A vector field that describes the magnetic effect on the moving electrical charges and magnetic materials as described above according to the induced emf and angular velocity. In a magnetic field, a charge that travels experiences a force perpendicular to its own velocity and to the magnetic field.
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