Question

A 2.2 m -long wire carries a current of 7.2 A and is immersed within a uniform magnetic field B⃗ . When this wire lies along the +x axis, a magnetic force F⃗ =(−2.6j)N acts on the wire, and when it lies on the +y axis, the force is F⃗ =(2.6i−4.4k)N.

Answer 1

Answer:

Explanation:

Given that :

The current I =  7.2 A

The length vector L =   2.2 m i  

The force vector is = -2.6 j N

When L = 2.2 m i ; the

Force vector F = (2.6 i - 4.4 k) N

Compute the components of the magnetic field as follows:

$F = I(L*B)$

Replacing 7.2 A for I ; -2.6 j N for F & 2.2 m i for L

$-2.6 j N = (7.2 A)(2.2 m i *(B_xi + B_yj+B_zk)$

$-2.6 j N = 15.84B_z \ and \ 0 = 15.84B_y$

$B_z = -0.1641T \ and \ B_y = 0$

However in y direction ; we have :

$(2.6 i - 4.4 k) = 7.2 A (2.2 mi*(B_xi+B_yj+B_zk)$

$- 2.6 = 15.84 B_z \ and \ -4.4 = 15.84 B_x$

$B_z = -0.1641 T \ and \ B_x = -0.2778T$

Hence, the component of magnetic field is as follows:

$B_x = -.02778T \ ; B_y = 0 T \ ; B_z = - 0.1641 T)$