The wires which connect the battery of an automobile to its starting motor carry a current of 300 A (for a short time). What is the force per unit length between the wires if they are 70cm long and 1.5cm apart? Is the force attractive or repulsive?
Answer
609.3k+ views
Hint
Use the formula for the force per unit length between the two wires as $F = \dfrac{{{\mu _0}{I^2}}}{{2\pi d}}$ where ${\mu _0}$ is the permeability of free space, $I$ is the current that flows through the wires and $d$ is the distance between the two wires. And, since both wires have the same source, current flows in the opposite direction.
Complete step-by-step answer
We know that the force per unit length is given by,
$ \Rightarrow f = {f_{ba}} = \dfrac{{{\mu _0}{I_a}{I_b}}}{{2\pi d}}$ Where, ${\mu _0}$ is the permeability of free space, ${I_a},{I_b}$ are the current through the two wires – a and b, and $d$ is the distance between the wires.
We substitute the values of ${\mu _0} = 4\pi \times {10^{ - 7}}Tm{A^{ - 1}}$ , ${I_a} = {I_b} = 300A$ and $d = 1.5cm = 1.5 \times {10^{ - 2}}m$ in the formula above and get,
$ \Rightarrow f = \dfrac{{4\pi \times {{10}^{ - 7}} \times 300 \times 300}}{{2\pi \times 1.5 \times {{10}^{ - 2}}}}N/m$
Cancelling $\pi $ from both the numerator and denominator and combining the powers of 10 together, we get,
$ \Rightarrow f = \dfrac{{4 \times 3 \times 3 \times {{10}^{4 - 7 + 2}}}}{{2 \times 1.5}}N/m$
Simplifying further we obtain,
$ \Rightarrow f = \dfrac{{18}}{{15}} = 1.2N/m$.
Now, we know that both wires are going from battery to the starter motor forming the part of a single circuit with them connected to the opposite ends of the battery, therefore, the direction of current in the two wires is opposite to each other. Hence, the force between the two wires is repulsive to each other since we know that parallel currents attract and anti-parallel currents repel each other.
Note
The question provides data for the length of the wires which is not required and is probably given to confuse the students with the distance between the two wires. This provides a slight scope for error.
Use the formula for the force per unit length between the two wires as $F = \dfrac{{{\mu _0}{I^2}}}{{2\pi d}}$ where ${\mu _0}$ is the permeability of free space, $I$ is the current that flows through the wires and $d$ is the distance between the two wires. And, since both wires have the same source, current flows in the opposite direction.
Complete step-by-step answer
We know that the force per unit length is given by,
$ \Rightarrow f = {f_{ba}} = \dfrac{{{\mu _0}{I_a}{I_b}}}{{2\pi d}}$ Where, ${\mu _0}$ is the permeability of free space, ${I_a},{I_b}$ are the current through the two wires – a and b, and $d$ is the distance between the wires.
We substitute the values of ${\mu _0} = 4\pi \times {10^{ - 7}}Tm{A^{ - 1}}$ , ${I_a} = {I_b} = 300A$ and $d = 1.5cm = 1.5 \times {10^{ - 2}}m$ in the formula above and get,
$ \Rightarrow f = \dfrac{{4\pi \times {{10}^{ - 7}} \times 300 \times 300}}{{2\pi \times 1.5 \times {{10}^{ - 2}}}}N/m$
Cancelling $\pi $ from both the numerator and denominator and combining the powers of 10 together, we get,
$ \Rightarrow f = \dfrac{{4 \times 3 \times 3 \times {{10}^{4 - 7 + 2}}}}{{2 \times 1.5}}N/m$
Simplifying further we obtain,
$ \Rightarrow f = \dfrac{{18}}{{15}} = 1.2N/m$.
Now, we know that both wires are going from battery to the starter motor forming the part of a single circuit with them connected to the opposite ends of the battery, therefore, the direction of current in the two wires is opposite to each other. Hence, the force between the two wires is repulsive to each other since we know that parallel currents attract and anti-parallel currents repel each other.
Note
The question provides data for the length of the wires which is not required and is probably given to confuse the students with the distance between the two wires. This provides a slight scope for error.
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