Calculate the magnitude of force on a straight wire, carrying current, when placed in a magnetic field and direction of magnetic force on a ring of radius, carrying current, placed in a magnetic field.

- Current carrying wire experiences a force when placed in a magnetic field.
- The force experienced by a wire is the vector sum of individual force exerted on all charges to form a current.
- Consider a wire carrying current \(I\) placed in magnetic field as shown in figure.

- By right hand thumb rule, the force on a current carrying wire due to magnetic field will be at left side.

- Magnetic force on a charge q, moving with drift velocity v
_{d}in a magnetic field B is

\(\vec F = q (\vec v_d × \vec B)\)

- The magnetic force on a segment of wire of length L and cross
^{_}sectional area is given as

\(\vec F = q (\vec V_d × \vec B) n \,A\,L\)

where, n is the number of charge carrier per unit volume.

- Relation between current and drift velocity is given as

\(I = n\,q \, v_d \,A\)

- Thus magnetic force on wire,

\(\vec F = I (\vec L × \vec B)\)

- Direction of \(\vec L\) is taken in the direction of current and \(|\vec L| = L\) ,i.e., length of segment.

- Consider a wire of length \(L\), placed in a uniform magnetic field \(\vec B\), making an angle \(\theta\) with field.

- Magnetic force on a current carrying wire is given as

\(\vec F_B =\text I (\vec L × \vec B )\)

\(|\vec F_B| =\text{ I L B}\, sin\, \theta\)

where,

\(\text L\,sin \,\theta\) is a perpendicular length

\(\text I\) is current

\(B\) is magnetic field

- Consider a wire of arbitrary shape, placed in a uniform magnetic field.

- To calculate force on the wire placed in a magnetic field, choose a small element \({d\vec\ell}\) and analyze force on it.
- The direction of force on this element is outside the page by using right hand thumb rule.
- Force on element \(d\ell\)

\((\vec F_B)_{d\ell} = I ( {d\vec\ell} × \vec B) \)

- So, the force on whole wire is

\(\vec F_B = I\displaystyle \int_P^Q\, {d\vec\ell} × \vec B \)

\(\vec F_B = I (\vec {PQ}) × \vec B\)

where, P and Q represent the end point of the wire.

A \(6 \, \hat k \, N\)

B \(8 \, \hat k \, N\)

C \(12 \, \hat k \, N\)

D \(10 \, \hat k \, N\)

- Consider a rod of length \( L\) and mass m carrying current \(\text I\), is placed on a smooth surface in a uniform magnetic field.

- Force on a rod due to magnetic field

\(F = ILB\) ..........(1)

\(\Big[\)Since, \(\vec B\) is perpendicular to \(\vec L\) so, \(\vec B × \vec L = BL \Big]\)

- By free body diagram of rod

Along y-axis,

N = mg [Normal force cancels gravitational force]

So, \(\Sigma F = ma \)

F = ma ..........(2)

On comparing (1) and (2), we get

\(I L B =ma\)

\(a = \dfrac{ILB}{m}\)

- Consider a sine shaped current ( \(\text I\)) carrying wire, placed in uniform magnetic field (B).

- The length of sine shaped wire

\(\vec{PQ} = 4R\)

- Hence, force on current carrying sine shaped wire
\(\vec F = I (\vec{PQ }× \vec B)\) \(\Big[\vec {PQ} = 4R\Big]\)

\(\vec F = I \,4 R\,B\, \hat k \,\) (outside)

where,

\(\text I \) is current

\(R\) is radius of wire loop

\(B\) is magnetic field

A \(8\,\hat k\,N\)

B \(16\,\hat k\,N\)

C \(5\,\hat k\,N\)

D \(4\,\hat k\,N\)

- Consider a strong magnet, placed under a horizontal conducting ring of radius r, carrying current \(\text I\) as shown in figure.
- A magnetic field B makes an angle \(\theta\) with the vertical at the ring location.

Here, r is radius of ring and \(\text I\) is the current.

- Horizontal components of force will cancel out and only the perpendicular components get added.

- Thus, net force will be in vertical direction.

\(\vec F_{d\ell}=\text I \,d\ell\,\text {B sin}\,\theta\)

\(\vec F = \displaystyle \int \limits _0^{2\pi r} I d\,\ell \, B \, sin\,\theta\)

\(\vec F = I \, B \, sin\,\theta\displaystyle \int \limits _0^{2\pi r} d\,\ell \)

\(\vec F = I B \,sin\, \theta × 2 \pi \,r\)

\(\vec F = 2 \pi \,r \,I B \,sin\, \theta \) [upward direction ]

A \(8 \,\pi\, N \)

B \(6 \,\pi\, N \)

C \(16\,\pi\, N \)

D \(12\,\pi\, N \)