
A straight wire of length 0.5 metre and carrying a current of 1.2 ampere is placed in a uniform magnetic field of induction 2 tesla. The magnetic field is perpendicular to the length of the wire.
The force on the wire is
A. $2.4$
B. $1.2$
C. $2.0$
D. $3.0$
Answer
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Hint:We will use the general formula of magnetic force to calculate the magnitude of force based on the given value of parameters. In magnetism, when a current carrying wire is placed in a magnetic field, it experiences a magnetic force that depends on the magnitude of current, the length of the wire, and the magnetic field in which it is placed.
Formula Used:
$F = i\vec L \times \vec B$
Complete answer:
Electromagnetic force is a function of charge motion, which also produces magnetic force. In this situation, the magnetic force can be thought of as a force that results from magnetic fields interacting.
According to the right-hand rule, the conductor will experience force in the direction indicated by the thumb if the index finger points in the direction of the current flowing through the wire and the middle finger points in the direction of the magnetic field line at the location.
As we've mentioned in the problem,
$L = 0.5m$
$i = 1.2Amp$
$B = 2T$
A current-carrying wire in a magnetic field experiences force from,
$F = i\vec L \times \vec B$
Since the magnetic field is perpendicular to length of wire we can write
$F = iLB\sin 90$
$ \Rightarrow F = 1.2 \times 0.5 \times 2 = 1.2N$
Therefore, a magnetic force of magnitude will be experienced by the current-carrying wire while it is in a magnetic field. $F = 1.2N$
Hence option B is correct.
Thus, the correct option is B.
Note:It should be noted that when solving the problem, the fundamental unit of conversions—the Newton, represented by the letter N in the SI is used. Additionally, a wire carrying current produces its own magnetic field. As a result, when the wire is placed in another magnetic field, it will either be attracted or repellent according to the magnetic force present between the two fields.
Formula Used:
$F = i\vec L \times \vec B$
Complete answer:
Electromagnetic force is a function of charge motion, which also produces magnetic force. In this situation, the magnetic force can be thought of as a force that results from magnetic fields interacting.
According to the right-hand rule, the conductor will experience force in the direction indicated by the thumb if the index finger points in the direction of the current flowing through the wire and the middle finger points in the direction of the magnetic field line at the location.
As we've mentioned in the problem,
$L = 0.5m$
$i = 1.2Amp$
$B = 2T$
A current-carrying wire in a magnetic field experiences force from,
$F = i\vec L \times \vec B$
Since the magnetic field is perpendicular to length of wire we can write
$F = iLB\sin 90$
$ \Rightarrow F = 1.2 \times 0.5 \times 2 = 1.2N$
Therefore, a magnetic force of magnitude will be experienced by the current-carrying wire while it is in a magnetic field. $F = 1.2N$
Hence option B is correct.
Thus, the correct option is B.
Note:It should be noted that when solving the problem, the fundamental unit of conversions—the Newton, represented by the letter N in the SI is used. Additionally, a wire carrying current produces its own magnetic field. As a result, when the wire is placed in another magnetic field, it will either be attracted or repellent according to the magnetic force present between the two fields.
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