
A coil having $500$ square loops each of side $10\,cm$ is placed normal to a magnetic field which increases at the rate of $1\,T\,{s^{ - 1}}$ . The induced e.m.f. is
A. $0.1\,V$
B. $5.0\,V$
C. $0.5\,V$
D. $1.0\,V$
Answer
557.7k+ views
Hint: the E.M.F. is induced in the coil when the flux in a conductor or coil changes. Here we will first change the length of the loop from meter into centimeter. Also, here we will use the formula of induced E.M.F in terms of magnetic field.
Formula used:
The formula of induced E.M.F is given by
$E.M.F = \dfrac{d}{{dt}}\left( {NBA} \right)$
Here, $N$ is the number of turns of the coil, $B$ is the magnetic field and $A$ is the area of the loop.
Complete step by step answer:
Consider a coil having $500$ square loops. These square loops is of $10\,cm$ . Let these loops are placed perpendicular to a magnetic field.
Therefore, the number of turns are $N = 500$
Also, the side of the square loop is $l = 10\,cm$
Now, changing the length of the loop from meters into centimeters as given below
$l = 10\,cm = \dfrac{{10}}{{100}}m$
$ \Rightarrow \,l = 0.1\,m$
Also, the rate of change of magnetic is given by
$\dfrac{{dB}}{{dt}} = 1\,T\,{s^{ - 1}}$
Also, the area of the square loops is given by
$A = {\left( {0.1} \right)^2}$
$ \Rightarrow \,A = 0.01\,{m^2}$
An E.M.F. or electromotive force is defined as the electric potential that is produced by changing the magnetic field. Therefore, an electromotive force is induced when the flux in a conductor or coil changes. Therefore, the E.M.F. induced in the coil is given by
$E.M.F = \dfrac{d}{{dt}}\left( {NBA} \right)$
$ \Rightarrow \,E.M.F. = NA\dfrac{{dB}}{{dt}}$
Now, putting the value of the rate of the change of magnetic field, we get
$ \Rightarrow E.M.F. = NA$
$ \Rightarrow \,E.M.F. = 500 \times 0.01$
$ \Rightarrow \,E.M.F = 5.0\,V$
Therefore, the E.M.F induced in the coil having $500$ square loops is $5.0\,V$ .
So, the correct answer is “Option B”.
Note:
The induced E.M.F is also known as induced electromotive force, electromagnetic induction, and electromotive force. Here, the E.M.F. is induced by the magnetic field of the oil whose loops are placed at normal to the magnetic field. The E.M.F. induced in the coil generates the potential difference across the coil.
Formula used:
The formula of induced E.M.F is given by
$E.M.F = \dfrac{d}{{dt}}\left( {NBA} \right)$
Here, $N$ is the number of turns of the coil, $B$ is the magnetic field and $A$ is the area of the loop.
Complete step by step answer:
Consider a coil having $500$ square loops. These square loops is of $10\,cm$ . Let these loops are placed perpendicular to a magnetic field.
Therefore, the number of turns are $N = 500$
Also, the side of the square loop is $l = 10\,cm$
Now, changing the length of the loop from meters into centimeters as given below
$l = 10\,cm = \dfrac{{10}}{{100}}m$
$ \Rightarrow \,l = 0.1\,m$
Also, the rate of change of magnetic is given by
$\dfrac{{dB}}{{dt}} = 1\,T\,{s^{ - 1}}$
Also, the area of the square loops is given by
$A = {\left( {0.1} \right)^2}$
$ \Rightarrow \,A = 0.01\,{m^2}$
An E.M.F. or electromotive force is defined as the electric potential that is produced by changing the magnetic field. Therefore, an electromotive force is induced when the flux in a conductor or coil changes. Therefore, the E.M.F. induced in the coil is given by
$E.M.F = \dfrac{d}{{dt}}\left( {NBA} \right)$
$ \Rightarrow \,E.M.F. = NA\dfrac{{dB}}{{dt}}$
Now, putting the value of the rate of the change of magnetic field, we get
$ \Rightarrow E.M.F. = NA$
$ \Rightarrow \,E.M.F. = 500 \times 0.01$
$ \Rightarrow \,E.M.F = 5.0\,V$
Therefore, the E.M.F induced in the coil having $500$ square loops is $5.0\,V$ .
So, the correct answer is “Option B”.
Note:
The induced E.M.F is also known as induced electromotive force, electromagnetic induction, and electromotive force. Here, the E.M.F. is induced by the magnetic field of the oil whose loops are placed at normal to the magnetic field. The E.M.F. induced in the coil generates the potential difference across the coil.
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