
3 A of current is flowing in a linear conductor having a length of 40 cm. The conductor is placed in a magnetic field of strength 500 gauss and makes an angle of ${30^0}$ with direction of the field. It experiences a force of magnitude
a) $3 \times {10^4}N$
b) $3 \times {10^2}N$
c) $3 \times {10^{ - 2}}N$
d) $3 \times {10^{ - 4}}N$
Answer
164.1k+ views
Hint: When current flows on a straight conductor of a specific length carrying current and if placed in a magnetic field, it experience a force of a definite magnitude. So by putting given values in the general relation of force $F$ on the current carrying conductor we can find out the required result.
Formula used:
A general relation of force on the current carrying conductor
$F = ilB\sin \theta $
Where,
$F=$ Magnetic force on a current
$i\,\,\And \,\,B$ represents flowing current and magnetic field.
$l=$ Length of the conductor
$\theta =$ Angle between current element vector and magnetic field vector
Complete answer:
When a current travels through a conductor in a magnetic field, a force is exerted on the conductor. This force is called the Lorentz force. The magnitude of the force is proportional to the current and the strength of the magnetic field. The direction of the force is ${90^o}$ from both the current and the magnetic field.
The formula force on a current carrying conductor to this magnetic field is:
$F = ilB\sin \theta $
Where, $i$ is the current flowing through the conductor, $l$ is the length of the conductor, $B$ is the magnetic field and $\theta $=angle between the current element vector and magnetic field.
Firstly we have to convert the magnetic field strength in the S.I unit too. Hence if we convert it we get \[1{\text{ }}gauss = {10^{ - 4}}tesla\].
Now we can easily find out the magnitude of force using the formula
$F = ilB\sin \theta $
Or, $F = 3 \times 40 \times {10^{ - 2}} \times 500 \times {10^{ - 4}} \times \sin {30^0} = 3 \times {10^{ - 4}}N$
So the magnitude of force is $3 \times {10^{ - 4}}N$.
Hence the correct option is (D).
Note: While solving such numericals, always make sure that all the given parameters values are in the same units or not if not always convert the values in the same unit first and then put all the values in order to avoid any calculation mistakes.
Formula used:
A general relation of force on the current carrying conductor
$F = ilB\sin \theta $
Where,
$F=$ Magnetic force on a current
$i\,\,\And \,\,B$ represents flowing current and magnetic field.
$l=$ Length of the conductor
$\theta =$ Angle between current element vector and magnetic field vector
Complete answer:
When a current travels through a conductor in a magnetic field, a force is exerted on the conductor. This force is called the Lorentz force. The magnitude of the force is proportional to the current and the strength of the magnetic field. The direction of the force is ${90^o}$ from both the current and the magnetic field.
The formula force on a current carrying conductor to this magnetic field is:
$F = ilB\sin \theta $
Where, $i$ is the current flowing through the conductor, $l$ is the length of the conductor, $B$ is the magnetic field and $\theta $=angle between the current element vector and magnetic field.
Firstly we have to convert the magnetic field strength in the S.I unit too. Hence if we convert it we get \[1{\text{ }}gauss = {10^{ - 4}}tesla\].
Now we can easily find out the magnitude of force using the formula
$F = ilB\sin \theta $
Or, $F = 3 \times 40 \times {10^{ - 2}} \times 500 \times {10^{ - 4}} \times \sin {30^0} = 3 \times {10^{ - 4}}N$
So the magnitude of force is $3 \times {10^{ - 4}}N$.
Hence the correct option is (D).
Note: While solving such numericals, always make sure that all the given parameters values are in the same units or not if not always convert the values in the same unit first and then put all the values in order to avoid any calculation mistakes.
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