Question

The right edge of the circuit in the figure extends into a 50 mT uniform magnetic field.

What are the magnitude and direction of the net force on the circuit?

	left
	right
	up
	down

Answer 1

Concepts and reason

The concepts required to solve the problem are Lorentz magnetic force, induced current, Ohm’s law and Fleming’s left-hand rule.

Determine the induced current in the circuit due to the source voltage using Ohm’s law. Use the current, magnetic field and the length of the loop to calculate the magnetic force. Use the left hand rule to determine the direction of the force.

Fundamentals

The Lorentz magnetic force is,

$F = ILB\sin \theta$

Here, $I$ is the current, $l$ is the length of the wire, $B$ is the magnetic field, and $\theta$ is the angle between the direction of current, and the magnetic field.

According to Ohm’s law, the current through a resistor due to a source voltage is,

$I = \frac{V}{R}$

Here, $V$ is the voltage of the source and $R$ is the resistance.

Flemings left hand states “the middle finger represents the direction of the current, index finger represents direction of the magnetic field, and the thumb represents the direction of the force.

According to Ohm’s law, the current through the circuit is,

$I = \frac{V}{R}$

Substitute $15\,{\rm{V}}$ for $V$ and $3.0\,\Omega$ for $R$in the equation $I = \frac{V}{R}$.

$\begin{array}{c}\\I = \frac{{15\,{\rm{V}}}}{{{\rm{3}}{\rm{.0}}\,\Omega }}\\\\ = 5.0\,{\rm{A}}\\\end{array}$

There are three portions of the circuit which is subjected to magnetic field. The top portion of the circuit and the lower portion of the circuit will experience same magnitude of magnetic force but opposite in direction. These fields get cancelled. Only the side portion will experience the effective magnetic force. The side is perpendicular to the magnetic field.

The Lorentz magnetic force is,

$F = ILB\sin \theta$

Substitute $5.0\,{\rm{A}}$ for $I$, $10\,{\rm{cm}}$ for $L$, $50\,{\rm{mT}}$ for $B$, and $90^\circ$ for $\theta$ in the equation $F = ILB\sin \theta$. The Lorentz magnetic force experienced on the side is,

$\begin{array}{c}\\F = \left( {5.0\,{\rm{A}}} \right)\left( {\left( {10\,{\rm{cm}}} \right)\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1\,{\rm{cm}}}}} \right)} \right)\left( {\left( {50\,{\rm{mT}}} \right)\left( {\frac{{{{10}^{ - 3}}{\rm{ T}}}}{{{\rm{1 mT}}}}} \right)} \right)\left( {\sin 90^\circ } \right)\\\\ = 2.5 \times {10^{ - 2}}\,{\rm{N}}\\\end{array}$

The magnitude of the net force on the circuit is $2.5 \times {10^{ - 2}}\,{\rm{N}}$.

According to the left hand rule, the thumb of the left hand points towards the direction of force when the index finger points towards the direction of magnetic field and the middle finger points towards the direction of current.

Here, the direction of current in the circuit flows from top to bottom of the circuit, that is, it points towards the downward direction and the magnetic field points into the page.

If the index finger points into the page and the middle finger points downwards then the thumb points towards the right.

Thus, the direction of the magnetic force is towards right.

Ans:

The magnitude of the net force on the circuit is $2.5 \times {10^{ - 2}}\,{\rm{N}}$.

The direction of the magnetic force is towards right.