
A horizontal wire of length \[10cm\]and mass \[0.3g\]carries a current of \[5A\]. The magnitude of the magnetic field which can support the weight of the wire is \[\left( {g = 10m/{s^2}} \right)\] .
A. \[3 \times {10^{ - 3}}T\]
B. \[6 \times {10^{ - 3}}T\]
C. \[3 \times {10^{ - 4}}T\]
D. \[6 \times {10^{ - 4}}T\]
Answer
568.8k+ views
Hint: As we know that the force of a current carrying conductor will increase if the magnetic field strength increases, length of the conductor increases and Current passing from the conductor increases, As force is directly proportional to the magnetic field strength, current and the length of the conductor.
Formula used:
\[F = BIl\] ,
Here F is the force on the conductor, B is the magnetic field, I is the current flowing in the conductor and l is the current flowing from the conductor. As the magnitude of the magnetic field supports the weight of the wire, we find the equilibrium state by taking, \[F = mg\] .
Complete step by step answer:
Current passes through horizontal wire- \[5A\]
Length of the horizontal wire- \[10cm = 0.1m\]
Mass of the wire is \[0.3g = 0.3 \times {10^{ - 3}}\]
\[g = 10m/{s^2}\]
So, weight of the wire is \[ mg = 0.3 \times {10^{ - 3}} \times 10 = 0.003N\]
Assume B is the magnetic field strength applied to the horizontal wire.
As we know,
\[F = BIl\]
Here F is force, B is the magnetic field, I is current and l is the length of the wire.
\[F = B \times 0.1 \times 5\]
\[\Rightarrow F = 0.5B\]
In equilibrium,
We know,
\[F = mg\]
Now, equate these two things, we get-
\[0.5B = 0.003\]
\[\Rightarrow B = \dfrac{{0.003}}{{0.5}}\]
\[\therefore B = 0.006T\]
So, the magnetic field is \[6 \times {10^{ - 3}}T\].
Hence Option B is correct.
Note:Direction of magnetic field on a current carrying wire can be determined by right hand rule, which states that the direction of current represented by thumb and curling of finger represents the direction of magnetic field. If the thumb is in the upward direction, then the direction of the magnetic field will be anticlockwise and if the thumb is in downward direction then the magnetic field direction will be clockwise.
Formula used:
\[F = BIl\] ,
Here F is the force on the conductor, B is the magnetic field, I is the current flowing in the conductor and l is the current flowing from the conductor. As the magnitude of the magnetic field supports the weight of the wire, we find the equilibrium state by taking, \[F = mg\] .
Complete step by step answer:
Current passes through horizontal wire- \[5A\]
Length of the horizontal wire- \[10cm = 0.1m\]
Mass of the wire is \[0.3g = 0.3 \times {10^{ - 3}}\]
\[g = 10m/{s^2}\]
So, weight of the wire is \[ mg = 0.3 \times {10^{ - 3}} \times 10 = 0.003N\]
Assume B is the magnetic field strength applied to the horizontal wire.
As we know,
\[F = BIl\]
Here F is force, B is the magnetic field, I is current and l is the length of the wire.
\[F = B \times 0.1 \times 5\]
\[\Rightarrow F = 0.5B\]
In equilibrium,
We know,
\[F = mg\]
Now, equate these two things, we get-
\[0.5B = 0.003\]
\[\Rightarrow B = \dfrac{{0.003}}{{0.5}}\]
\[\therefore B = 0.006T\]
So, the magnetic field is \[6 \times {10^{ - 3}}T\].
Hence Option B is correct.
Note:Direction of magnetic field on a current carrying wire can be determined by right hand rule, which states that the direction of current represented by thumb and curling of finger represents the direction of magnetic field. If the thumb is in the upward direction, then the direction of the magnetic field will be anticlockwise and if the thumb is in downward direction then the magnetic field direction will be clockwise.
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