Question

A conducting rod of 1-meter length and 1 kg mass is suspended by two vertical wires through its end. An external magnetic field of 2 teslas is applied normally to the rod. Now, the current to be passed through the rod so as to make the tension in the wire is zero [ take \[\begin{array}{*{20}{c}} g& = &{10m{s^{ - 1}}} \end{array}\]].
A. 0.5 Amp
B. 15 Amp
C. 5 Amp
D. 1.5 Amp

Answer 1

Hint:Force-induced due to the magnetic field in the conducting rod will be equal to the force developed due to the gravitational acceleration. That whenever a current-carrying conductor is placed in a magnetic field, it is susceptible to a force provided by the magnetic field. In this case, magnetic Field Force on a Current-Carrying Conductor can be calculated using the below mentioned formula.

Formula used:
\[F = B \times i \times l\]……….. (a)
Where
\[F = \]magnetic force
\[B = \]Magnetic field
\[i = \]Current
\[l = \]length of the rod

Complete step by step solution:
In the question, we have given that there is a conducting rod whose length and mass \[1m\] and \[1{\rm{ }}kg\] respectively.
Mass of the rod (m) \[ = 1{\rm{ }}kg\], length of the rod (l) \[ = 1\]meter.
Magnetic field(B) \[ = 2\]Tesla, already given acceleration due to gravity (g) \[ = 10m{s^{ - 1}}\].
And we have to determine the current (i) passing through the rod.
Now we know that the magnetic force induced in the conducting rod due to the magnetic field is,
⟹\[F = B \times i \times l\] ……….. (a)

Now, force developed in the rod due to the gravitational acceleration is,
⟹\[F = mg\]……….. (b)
Where
mass of the rod.
Now from the equation (a) and (b), we can write,
⟹\[mg = B \times i \times l\]
Now put the given values, so
⟹\[1 \times 10 = 2 \times i \times 1\]
⟹\[i = 5\,A\]
Now the final answer is \[5\,A\].

Therefore, the correct option is C.

Note: Remember that If there is acceleration due to gravity, then the developed force will always be \[F = ma\]. Students should remember that when the current carrying conductors is parallel to the field, the applying force must be zero. It is also important to keep in mind that magnetic force operating on a current carrying conductor is not a central force because the expression does not depend upon r.