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# An electric cell does 5 joules of work in carrying 10-coulomb charge around the closed circuit. The electromotive force of the cell is:A. 2 voltB. 0.5 voltC. 4 voltD. 1 volt.

Last updated date: 10th Sep 2024
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Answer
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Hint:-The electromotive force or the e.m.f is the energy supplied to the charge passing in the circuit and electromotive force or the e.m.f is the potential difference of the battery on the terminals of the cell when no current is flowing from the circuit.
Formula used: The formula of the electromotive cell is given by,
$e = \dfrac{W}{Q}$
Where e is the electromotive force or the e.m.f and E is the work done to move charge and Q is the charge in coulomb also the unit of electromagnetic force or e.m.f is volts.

Complete step-by-step solution
It is given that a cell does a 5 joule of work for carrying a charge of 10 coulomb in a closed circuit and we need to find the value of the electromagnetic force or e.m.f
As the formula of the electromagnetic force is given by,
$e = \dfrac{W}{Q}$
Where e is the electromotive force or the e.m.f and E is the work done to move charge and Q is the charge in coulomb also the unit of electromagnetic force or e.m.f is volts.
Replace the value of the work done and the charge in the above relation,
$\Rightarrow e = \dfrac{W}{Q}$
Put $W = 5J$ and $Q = 10C$ in the above relation.
$\Rightarrow e = \dfrac{5}{{10}}$
$\Rightarrow e = \dfrac{1}{2}V$
So the electromagnetic force or e.m.f for this problem is $e = \dfrac{1}{2}V$ where V is volts. The correct answer for this problem is option B.

Note:- The e.m.f is the potential difference of the battery along the terminal when there is no current flowing in the circuit. The electromagnetic force is the work required to move a unit charge from one terminal to another terminal following the circuit.